| Skład początkowy: Ołów surowey |
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Nazwa pierwiastka |
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Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
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Symbol pierwiastka |
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Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
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Udział procentowy [%] |
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0,18 |
0,3 |
0,25 |
0,26 |
0,09 |
0,59 |
98 |
0,21 |
0,12 |
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Udział masowy [kg] |
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1,8 |
3 |
2,5 |
2,6 |
0,9 |
5,9 |
980 |
2,1 |
1,2 |
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| I ETAP: Usuwanie miedzi |
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1. Odmiedziowanie segregacyjne |
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Na 1000kg ołowiu surowego przypada (masowo): |
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(procentowo) |
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Pb |
980 |
kg |
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100% |
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Cu |
5,9 |
kg |
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x% |
0,602 |
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| Z układu równowagii Cu- Pb odczytałem, że taki stop w po obniżeniu temp. z 550°C do 326°C będzie zawierał monotektyczną fazę ciekłą i czystą miedź. |
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Znam, skład stopu Cu-Pb: |
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W układzie fazowym skład fazy ciekłej wygląda następująco |
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Pb |
99,398 |
[%] |
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Pb |
99,94 |
% |
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Cu |
0,602 |
[%] |
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Cu |
0,060000000000002 |
% |
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| Ilość miedzi pozostałej w roztworze |
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0,602 |
% |
100% |
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0,060000000000002 |
% |
x% |
9,97% |
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| Miedź wydzielona w postaci kryształów (procentowo) |
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100% - 9,97% = |
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90,03% |
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100% |
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| Ilość wydzielonych kryształów (masowo) |
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100 |
% |
5,9 |
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90,03 |
% |
x |
5,31177 |
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| Ilość miedzi pozostała w roztowrze (masowo) |
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5,9 |
-5,31177 |
= |
0,58823 |
kg |
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| Do szklirów oprócz miedzi i ołowiu przechodzi arsen i antymon |
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Na 1000kg ołowiu surowego przypada (masowo): |
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(procentowo) |
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Pb |
980 |
kg |
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100% |
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As |
3 |
kg |
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x% |
0,306 |
% |
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| Z układu równowagii As- Pb odczytałem, że taki stop w po obniżeniu temp. z 550°C do 326°C będzie zawierał fazę ciekłą i czysty ołów. |
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Znam, skład stopu As-Pb: |
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W układzie fazowym skład fazy ciekłej wygląda następująco |
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Pb |
99,694 |
[%] |
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Pb |
25 |
% |
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As |
0,306 |
[%] |
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As |
75 |
% |
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| Obliczam, ile As się wydzieli: |
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99,694 |
Pb% |
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0,306 |
As% |
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25 |
Pb% |
- |
75 |
As% |
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As |
3 |
kg |
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100 |
As% |
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3 |
As kg |
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75 |
As% |
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x |
As kg |
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x |
2,250 |
As kg |
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| Dodatkowo z arsenem wydzieli się ołów |
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75 |
As % |
= |
2,250 |
kg |
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25 |
Pb % |
= |
x |
kg |
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x |
0,75 |
Pb kg |
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| Ilość arsenu i ołowiu pozostała w roztworze (masowo) |
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3 |
- |
2,250 |
= |
0,750 |
kg |
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980 |
- |
0,75 |
= |
979,25 |
kg |
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Na 1000kg ołowiu surowego przypada (masowo): |
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(procentowo) |
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Pb |
980 |
kg |
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100% |
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Sb |
1,8 |
kg |
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x% |
0,184 |
% |
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| Z układu równowagii Sb- Pb odczytałęm, że taki stop w po obnizeniu temp. z 550°C do 326°C będzie zawierał fazę ciekłą i czysty ołów. |
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Znamy więc, skład stopu sb-Pb: |
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W układzie fazowym skład fazy ciekłej wygląda następująco |
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Pb |
99,816 |
[%] |
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Pb |
96 |
% |
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Sb |
0,184 |
[%] |
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Sb |
4 |
% |
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| Obliczam, ile Sb się wydzieli: |
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99,816 |
Pb% |
- |
0,184 |
Sb% |
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96 |
Pb% |
- |
4 |
Sb% |
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Sb |
1,8 |
kg |
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100 |
Sb% |
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1,8 |
Sb kg |
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4 |
Sb% |
= |
x |
Sb kg |
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x |
0,072 |
Sb kg |
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| Dodatkowo z arsenem wydzieli się ołów |
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4 |
Sb % |
= |
0,072 |
kg |
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96 |
Pb % |
= |
x |
kg |
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x |
1,728 |
Pb kg |
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| Ilość arsenu i ołowiu pozostała w roztworze (masowo) |
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1,8 |
- |
0,072 |
= |
1,728 |
kg |
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979,25 |
- |
1,728 |
= |
977,522 |
kg |
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| Głównym składnikiem szklirów jest ołów, ponieważ przy zdejmowaniu szklirów z powierzchni kąpieli dużymi dziurkowanymi łopatami, łopata zabiera razem z cząstkami stałymi dużo ołowiu, |
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| którego część nie zdąży spłynąć z powrotem przez jej dziurkowane dno do kąpieli (8,973 kg). |
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| Skład szklirów: |
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[kg] |
[%] |
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Ołów |
11,451 |
60 |
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Miedź |
5,312 |
27,83 |
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Arsen |
2,250 |
11,79 |
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Antymon |
0,072 |
0,38 |
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Razem |
100 |
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| Skład ołowiu po odmiedziowaniu segregacyjnym |
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Nazwa pierwiastka |
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Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
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Symbol pierwiastka |
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Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
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Udział procentowy [%] |
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0,176162052777254 |
0,076459224295683 |
0,25486408098561 |
0,265058644225035 |
0,09175106915482 |
0,059944031847816 |
98,7393403098128 |
0,214085828027913 |
0,122334758873093 |
Razem |
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Udział masowy [kg] |
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1,728 |
0,750 |
2,5 |
2,6 |
0,9 |
0,588 |
968,549 |
2,1 |
1,2 |
980,915 |
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| I ETAP: Usuwanie miedzi |
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2. Odmiedziowanie siarką |
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Siarczek miedziawy nie rozpuszcza się w Pb i w stanie stalym wypływa a powierzchnię , z której jest zdejmowany jako szlikry siarczkowe, |
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zawierające do 95% Pb i 3% Cu. Po odmiedziowaniu siarką zawartośc Cu w Pb zmniejsza się do 0,001%. |
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| Siarkę do kąpieli dodaję w stosunku |
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Reakcje chemiczne: |
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Pb + S = PbS |
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1 |
kg S |
- |
1000 |
kg Pb |
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Cu + PbS = CuS + Pb |
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x |
kg S |
- |
968,549 |
kg Pb |
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x= |
0,968549 |
kg S |
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PbS + 2Cu = Pb + Cu2S |
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szlikry: |
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Mam daną ilość Cu ołowiu rafinowanego (wg normy) |
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Po etapie Ia skład Cu w ołowiu surowym wynosi: |
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PbS * CuS |
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Kamienie Pb |
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Cu |
0,002 |
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Cu |
0,588 |
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Obliczam ile Cu przereagowało: |
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Obliczam wspólczynnik, mówiacy, ile razy dany pierwiastek jest lżejszy od Cu. Dzięki temu mogę ustalić wagę poszczególnych składnikow reakcji. |
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[kg] |
[unit] |
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pierwiastek |
[unit] |
Cu |
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Cu |
0,588 |
68,55 |
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Pb |
207,2 |
0,33 |
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S |
32,07 |
2,14 |
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razy lzejszy |
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Pb [kg] |
+ |
S [kg] |
= |
PbS [kg] |
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1,777 |
+ |
0,275 |
= |
1,50 |
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|
|
|
|
|
Cu [kg] |
+ |
PbS [kg] |
= |
CuS [kg] |
+ |
Pb [kg] |
|
|
|
|
|
|
|
|
|
|
0,194 |
+ |
0,68 |
= |
0,285 |
+ |
0,587 |
|
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|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
2Cu [kg] |
+ |
PbS [kg] |
= |
Cu2S [kg] |
+ |
Pb [kg] |
|
|
|
|
|
|
|
|
|
|
0,388 |
+ |
1,35 |
= |
0,57 |
+ |
1,17 |
|
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|
|
PbS [kg] |
+ |
CuS [kg] |
= |
PbS |
* |
CuS [kg] |
|
|
|
|
|
|
|
|
|
|
0,285 |
+ |
0,285 |
= |
|
0,570 |
|
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|
PbS [kg] pocz. |
- |
PbS[kg] zużyte |
= |
PbS [kg] |
|
|
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|
|
|
|
|
|
2,03 |
- |
0,28 |
= |
1,747 |
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|
Skład szlikrów: |
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|
|
Wystarczy więc dodać: |
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|
|
[kg] |
[%] |
|
|
S |
0,638 |
[kg] |
|
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|
|
|
|
|
|
Cu |
0,582 |
18 |
|
|
a nie: |
|
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|
|
|
|
|
|
|
|
|
S |
0,638200017505471 |
20 |
|
|
S |
0,969 |
[kg] |
|
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|
|
|
|
|
|
Pb |
1,941 |
62 |
|
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|
razem: |
3,161 |
100 |
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| Skład ołowiu po odmiedziowaniu segregacyjnym i odmiedziowaniu siarką |
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|
Nazwa pierwiastka |
|
Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
|
|
|
|
|
|
Symbol pierwiastka |
|
Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
|
|
|
|
|
|
Udział procentowy [%] |
|
0,176616361549233 |
0,07665640692241 |
0,255521356408034 |
0,265742210664356 |
0,091987688306892 |
0,000600986230272 |
98,7955867994602 |
0,214637939382749 |
0,122650251075856 |
Razem |
|
|
|
|
|
Udział masowy [kg] |
|
1,728 |
0,750 |
2,5 |
2,6 |
0,9 |
0,006 |
966,608 |
2,1 |
1,2 |
978,392 |
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| II ETAP: Usuwanie cyny, arsenu i antymonu |
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Rafinacja metodą Harrisa, przy użyciu za pomocą soli NaNO3, NaOH |
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| Mam dany skład domieszek ołowiu rafinowanego (wg normy) |
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|
Reakcje chemiczne: |
|
2Pb + O2 = 2PbO |
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|
|
Sn |
As |
Sb |
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|
|
PbO + Na2O= Na2PbO2 |
|
|
|
0,001 |
0,001 |
0,005 |
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|
4PbO + 8NaOH = 4Na2PbO2 + 4H2O |
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|
2As + 5Na2PbO2 + 2H2O = 2Na3AsO4+4NaOH + 5Pb |
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|
| Po I Etapie skład wyżej wymienionych domieszek wynosi: |
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|
Sn + 2Na2PbO2 + H2O = Na2SnO3+2NaOH + 2Pb |
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|
Sn |
As |
Sb |
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|
2Sb + 5Na2PbO2 + 2H2O = 2Na3SbO4+4NaOH + 5Pb |
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|
|
2,6 |
0,750 |
1,728 |
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| Obliczam ile domieszk usunięto: |
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|
[kg] |
[unit] |
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| Sn |
2,599 |
118,71 |
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| As |
0,749 |
74,92 |
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| Sb |
1,723 |
121,75 |
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masa atm. |
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| Obliczam wspólczynnik, mówiacy, ile razy dany pierwiastek jest lżejszy od Sn. Dzięki temu mogę ustalić wagę poszczególnych składnikow reakcji. |
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|
pierwiastek |
[unit] |
Sn |
|
|
[kg] |
|
|
[kg] |
|
|
[kg] |
|
|
|
|
|
Na |
22,99 |
5 |
|
Na3AsO4 |
133,34 |
|
2As |
0,749 |
|
As |
0,3745 |
|
|
|
|
|
O |
16 |
7 |
|
Na2SnO3 |
212,69 |
|
Sn |
2,599 |
|
Sn |
2,599 |
|
|
|
|
|
Pb |
207,2 |
1 |
|
Na3SbO4 |
133,83 |
|
2Sb |
1,723 |
|
Sb |
0,8615 |
|
|
|
|
|
N |
14,01 |
8 |
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|
H |
1,01 |
118 |
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As |
74,92 |
2 |
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Sb |
121,175 |
1 |
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|
razy lzejszy |
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|
|
5Pb |
+ |
2NaNO3 |
= |
Na2O |
+ |
5PbO |
+ |
N2 |
|
|
|
|
|
|
|
|
83,60 |
+ |
13,70 |
= |
5,00 |
+ |
90,00 |
+ |
2,30 |
kg |
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| Powstało: |
5PbO |
90,00 |
[kg] |
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|
|
PbO |
+ |
Na2O |
= |
Na2PbO2 |
|
|
|
|
|
|
|
|
|
|
|
|
52,50 |
+ |
14,60 |
= |
67,10 |
kg |
|
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|
|
4PbO |
+ |
8NaOH |
= |
4Na2PbO2 |
+ |
4H2O |
|
|
|
|
|
|
|
|
|
|
37,50 |
+ |
13,40 |
= |
47,90 |
+ |
3,00 |
kg |
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|
| Powstało: |
Na2PbO2 |
67,10 |
kg |
oraz: |
H2O |
3,00 |
kg |
|
|
|
|
|
|
|
|
|
|
4Na2PbO2 |
47,90 |
kg |
|
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|
|
razem |
115,00 |
kg |
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|
|
2As |
+ |
5Na2PbO2 |
+ |
2H2O |
= |
2Na3AsO4 |
+ |
4NaOH |
+ |
5Pb |
|
|
|
|
|
|
0,75 |
+ |
47,90 |
+ |
1,20 |
= |
12,50 |
+ |
5,40 |
+ |
30,90 |
[kg] |
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|
|
Sn |
+ |
2Na2PbO2 |
+ |
H2O |
= |
Na2SnO3 |
+ |
2NaOH |
+ |
2Pb |
|
|
|
|
|
|
2,60 |
+ |
19,20 |
+ |
0,60 |
= |
7,10 |
+ |
2,70 |
+ |
12,50 |
[kg] |
|
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|
|
2Sb |
+ |
5Na2PbO2 |
+ |
2H2O |
= |
2Na3SbO4 |
+ |
4NaOH |
+ |
5Pb |
|
|
|
|
|
|
1,72 |
+ |
47,90 |
+ |
1,20 |
= |
11,30 |
+ |
5,40 |
+ |
34,12 |
[kg] |
|
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|
|
|
|
|
| Pierwiastki wydzielone: |
|
|
|
Dodajemy więc: |
|
|
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|
|
|
|
|
|
|
kg |
|
|
NaNO3 |
13,70 |
kg |
|
|
|
|
|
|
|
|
|
|
As |
0,75 |
|
|
NaOH |
13,40 |
kg |
|
|
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|
|
|
|
|
|
|
Sn |
2,60 |
|
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|
|
Sb |
1,72 |
|
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|
Na |
w całości |
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O |
w całości |
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|
H |
w całości |
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|
|
| Skład ołowiu po usunięciu antymonu, arsenu i cyny |
|
|
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|
|
Nazwa pierwiastka |
|
Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
|
|
|
|
|
|
Symbol pierwiastka |
|
Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
|
|
|
|
|
|
Udział procentowy [%] |
|
0,000821925689698 |
0,000102740711212 |
0,256851778030748 |
0,000102740711212 |
0,092466640091069 |
0,000616444267274 |
99,3099933834982 |
0,215755493545828 |
0,123288853454759 |
Razem |
|
|
|
|
|
Udział masowy [kg] |
|
0,008 |
0,001 |
2,5 |
0,001 |
0,9 |
0,006 |
966,608 |
2,1 |
1,2 |
973,324 |
|
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|
|
| III ETAP: Usuwanie srebra i złota |
|
|
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|
|
Proces odsrebrzania cynkiem Metodą Parkessa |
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|
|
| Mam dany skład domieszek ołowiu rafinowanego (wg normy) |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
Ag |
Au |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0,002 |
0,001 |
kg |
|
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|
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|
|
| Po II Etapie skład wyżej wymienionych domieszek wynosił |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Ag |
Au |
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
2,1 |
1,2 |
kg |
|
|
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|
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|
|
| Obliczam ile domieszk usunięto: |
|
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|
|
|
[kg] |
[unit] |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| Ag |
2,098 |
107,87 |
|
|
|
|
|
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|
|
|
|
|
| Au |
1,199 |
196,97 |
|
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|
|
masa.atm. |
|
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|
|
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|
|
| Obliczam wspólczynnik, mówiacy nam, ile razy dany pierwiastek jest lżejszy od Ag. Dzięki temu mogę ustalić wagę poszczególnych składnikow reakcji. |
|
|
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|
|
| pierwiastek |
[unit] |
Ag |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| Au |
196,97 |
0,5 |
|
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|
|
| Zn |
65,39 |
1,6 |
|
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|
| Srebro: |
Po obniżeniu temperarury powstaje piana srebra. |
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|
Zn z Ag tworzy związki o temp. topn. wyższej niż ołów. Tworzą się związki międzymetaliczne i "wyciągają" z Pb Srebro i Złoto. |
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|
3Ag |
+ |
3Zn |
= |
Ag3Zn3 |
|
|
|
|
|
|
|
|
|
|
|
|
1,26 |
+ |
0,76 |
= |
2,02 |
|
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|
|
2Ag |
+ |
5Zn |
= |
Ag2Zn5 |
|
|
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0,84 |
+ |
1,27 |
= |
2,11 |
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| Złóto |
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Au |
+ |
Zn |
= |
AuZn |
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0,24 |
+ |
0,15 |
= |
0,39 |
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Au |
+ |
3Zn |
= |
AuZn3 |
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0,24 |
+ |
0,44 |
= |
0,68 |
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3Au |
+ |
5Zn |
= |
Au3Zn5 |
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0,72 |
+ |
0,73 |
= |
1,45 |
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Na podstawie powyższych reakcji obliczam zużycie Zn. |
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Zn |
3,34 |
kg |
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Ponieważ Zn może się utleniać, a także wchodzić w reakcjie z innymi pierwiastkami, dodaję więcej Zn.(wiem, że zawartość Zn w Pb wzrośnie do 0,58%) |
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Zawartość Zn przed procesem: |
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Zawartość Zn po procesie odsrebrzania: |
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skład Pb surowego |
Zn |
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skład Pb surowego |
Zn |
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|
973,32 |
0,900 |
kg |
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973,32 |
x |
kg |
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|
100 |
0,092 |
% |
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100 |
0,58 |
% |
|
x |
5,65 |
Zn kg |
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Obliczam ogólne zużycie Zn. |
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Zn |
8,99 |
kg |
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Proces odsrebrzania podzielony jest na 3 części, w których odpowiednio dodaje się różne zawartości Zn. |
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Zn [ułamkowo] |
Zn [liczbowo] |
Zn [kg] |
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Część I |
2/3 |
0,67 |
5,99 |
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Część II |
1/4 |
0,25 |
2,25 |
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Część III |
1/12 |
0,08 |
0,75 |
Jeśli mamy dużą zawartość Ag w roztworze. |
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| Część I |
Piana, w której usuwa się 100% złota i 70% srebra. |
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Dodaję |
5,99 |
Zn [kg] |
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| Srebro: |
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100 |
% |
= |
2,10 |
Ag kg |
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3Ag |
+ |
3Zn |
= |
Ag3Zn3 |
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|
70 |
% |
= |
x |
Ag kg |
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0,08 |
+ |
0,05 |
= |
0,13 |
kg |
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x |
= |
1,47 |
Ag [kg] |
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2Ag |
+ |
5Zn |
= |
Ag2Zn5 |
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0,06 |
+ |
0,08 |
= |
0,14 |
kg |
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5 |
Ag |
= |
0,14 |
kg |
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3 |
Ag |
= |
0,08 |
kg |
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| Złoto : |
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2 |
Ag |
= |
0,06 |
kg |
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Au |
+ |
Zn |
= |
AuZn |
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1 |
Ag |
= |
0,03 |
kg |
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0,24 |
+ |
0,15 |
= |
0,39 |
kg |
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Au |
+ |
3Zn |
= |
AuZn3 |
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0,24 |
+ |
0,44 |
= |
0,68 |
kg |
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3Au |
+ |
5Zn |
= |
Au3Zn5 |
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0,72 |
+ |
0,73 |
= |
1,45 |
kg |
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| Część II |
Piana, w której znajduje się pozostałe 30% srebra. |
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Dodaję |
2,25 |
Zn [kg] |
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| Srebro: |
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100 |
% |
= |
2,10 |
Ag kg |
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30 |
% |
= |
x |
Ag kg |
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3Ag |
+ |
3Zn |
= |
Ag3Zn3 |
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0,04 |
+ |
0,02 |
= |
0,06 |
kg |
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x |
= |
0,63 |
Ag kg |
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2Ag |
+ |
5Zn |
= |
Ag2Zn5 |
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5 |
Ag |
= |
0,06 |
kg |
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0,02 |
+ |
0,04 |
= |
0,06 |
kg |
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3 |
Ag |
= |
0,04 |
kg |
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2 |
Ag |
= |
0,02 |
kg |
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| Ilość Zn dodana do Pb: |
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8,24 |
kg |
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1 |
Ag |
= |
0,01 |
kg |
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| Ilość Zn zużyta: |
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|
1,50 |
kg |
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| Ilość Zn pozostała w ołowiu po procesie: |
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|
6,74 |
kg |
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| Piana zawiera: |
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| Ag |
0,198 |
[kg] |
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| Au |
1,199 |
[kg] |
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| Skład ołowiu po usunięciu srebra i złota |
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Nazwa pierwiastka |
|
Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
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|
Symbol pierwiastka |
|
Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
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|
Udział procentowy [%] |
|
0,000819783843495 |
0,000102472980437 |
0,256182451092208 |
0,000102472980437 |
0,690667888144594 |
0,000614837882621 |
99,0512026741349 |
0,000204945960874 |
0,000102472980437 |
Razem |
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|
Udział masowy [kg] |
|
0,008 |
0,001 |
2,5 |
0,001 |
6,74 |
0,006 |
966,608 |
0,002 |
0,001 |
975,867 |
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| IV ETAP: Usuwanie cynku |
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Przy użyciu soli NaOH |
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Reakcje chemiczne |
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Zn + 2NaOH = Na2O*ZnO + H2 |
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Mam dany skład Zn w ołowiu rafinowanego (wg normy) |
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Zn |
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0,002 |
kg |
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Po III Etapie skład wyżej wymienionych domieszek wynosił |
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Zn |
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6,74 |
kg |
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Obliczay ile domieszk usunięto: |
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[kg] |
[unit] |
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Zn |
6,74 |
65,39 |
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Obliczam wspólczynnik, mówiacy nam, ile razy dany pierwiastek jest lżejszy od Zn. Dzięki temu mogę ustalić wagę poszczególnych składnikow reakcji. |
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pierwiastek |
[unit] |
Zn |
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Na |
22,99 |
2,8 |
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O |
16,00 |
4,1 |
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H |
1,01 |
64,7 |
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Reakcja zachodząca w procesie: |
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Zn |
+ |
2NaOH |
= |
Na2O*ZnO |
+ |
H2 |
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|
6,738 |
+ |
8,243 |
= |
14,773 |
+ |
0,208 |
kg |
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Powstałe związki, które się wydzielą, mają skład: |
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kg |
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Zn |
6,738 |
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Na |
w całości |
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O |
w całości |
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H |
w całości |
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Należy dodać: |
NaOH |
8,243 |
kg |
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| Skład ołowiu po usunięciu cynku |
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Nazwa pierwiastka |
|
Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
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Symbol pierwiastka |
|
Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
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|
Udział procentowy [%] |
|
0,000825483501164 |
0,000103185437646 |
0,25796359411389 |
0,000103185437646 |
0,000206370875291 |
0,000619112625873 |
99,7398695116956 |
0,000206370875291 |
0,000103185437646 |
Razem |
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|
Udział masowy [kg] |
|
0,008 |
0,001 |
2,5 |
0,001 |
0,002 |
0,006 |
966,608 |
0,002 |
0,001 |
969,129 |
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| V ETAP: Usuwanie bizmutu |
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Przy użyciu Ca i Mg |
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Mam dany skład Zn w ołowiu rafinowanego 99,97 (wg normy) |
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Bi |
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|
0,02 |
kg |
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Po etapie III skład tych domieszek w ołowiu surowym wynosi: |
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Bi |
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2,5 |
kg |
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Obliczam ile domieszk usunięto: |
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kg |
[unit] |
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Bi |
2,48 |
208,98 |
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Obliczam wspólczynnik, mówiacy nam, ile razy dany pierwiastek jest lżejszy od Bi. Dzięki temu mogę ustalić wagę poszczególnych składnikow reakcji. |
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pierwiastek |
[unit] |
Bi |
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Bi |
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Mg |
24,31 |
8,6 |
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0,49 |
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Ca |
40,08 |
5,2 |
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Reakcje zachodzące w procesie: |
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2Bi |
+ |
3Ca |
= |
Bi2Ca3 |
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0,98 |
+ |
0,28 |
= |
1,262 |
kg |
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3Bi |
+ |
Ca |
= |
Bi3Ca |
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1,47 |
+ |
0,094 |
= |
1,564 |
kg |
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6Mg |
+ |
3Bi3Ca |
= |
Bi2Ca3 |
+ |
2Bi2Mg3 |
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0,113 |
+ |
1,548 |
= |
0,416 |
+ |
0,760 |
kg |
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Ilość dodanych zwiazków: |
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Ca |
0,40 |
kg |
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Mg |
0,80 |
kg |
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Ca i Mg usuwa się w całości poprzez dodanie NaOH. |
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| Skład końcowy: Ołów rafinowany |
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Nazwa pierwiastka |
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Antymon |
Arsen |
Bizmut |
Cyna |
Cynk |
Miedź |
Ołów |
Srebro |
Złoto |
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Symbol pierwiastka |
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Sb |
As |
Bi |
Sn |
Zn |
Cu |
Pb |
Ag |
Au |
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Udział procentowy [%] |
|
0,000827601332024 |
0,000103450166503 |
0,002069003330061 |
0,000103450166503 |
0,000206900333006 |
0,000620700999018 |
99,9957585431734 |
0,000206900333006 |
0,000103450166503 |
Razem |
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Udział masowy [kg] |
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0,008 |
0,001 |
0,02 |
0,001 |
0,002 |
0,006 |
966,608 |
0,002 |
0,001 |
966,649 |
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