Answer:
7 grams of iron is produced from the reduction of 10.0 g of iron (III) oxide.
Explanation:
[tex]Fe_2O_3+3CO\rightarrow 3CO_2+2Fe[/tex]
Moles of iron (III) oxide :
[tex]\frac{10.0g}{160 g/mol}=0.0625 mol[/tex]
According to reaction, 1 mol of iron(III) oxide gives 2 moles of iron metal.
Then 0.0625 moles of iron(III) oxide will give:
[tex]\frac{2}{1}\times 0.0625 mol=0.125 mol[/tex] of iron metal.
Mass of 0.125 moles of iron metal.
[tex]0.125 mol\times 56 g/mol=7 g[/tex]
7 grams of iron is produced from the reduction of 10.0 g of iron (III) oxide.