M.Fe=56g/mol....din tabel
M,Fe2O3= (2x56gFe+3x16gO)/mol= 160g/mol
-calculez masa de oxid de fier pur
p=m,purx100/m,impur
85/100= m,pur/800g----------------->m,pur= 680g
- din ecuatia reactiei deduc masa de fier
2mol.........................4mol
3C+2Fe2o3----> 3CO2 +4Fe
2x160g..........................4x56g
680g...............................x
x= 680x2x56/160=...............gFe.......calculeaza!!!!!
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