ron metal is produced in a blast furnace by the reaction of iron (III) oxide and coke (pure carbon). If 25.0 moles of pure Fe2O3 is used, how many grams of iron can be produced?
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Expert's answer
2016-06-17T04:58:04-0400
Solution: n(Fe2O3)=25.0mol n(Fe0)-?
1. For calculation n(Fe) we need to calculate m(Fe2O3): Mr(Fe2O3)=55.85*2+16*3=159.6(g/mol) n=m/Mr; hereof m(Fe2O3)=n*Mr=159.6(g/mol)*25mol=3990g
2. Compose proportion: 3990(g) it is 159.6 X(g) it is 55.85*2 Where X is the total Iron weight in Fe2O3 Hereof X=3990*55.85*2/159.6=2792.5(g)
3. m(Fe0)=2792.5/55.85=50(mol) Answer: 50 mol of Iron can be produced from 25mol of Fe2O3
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