1) Balance the chemical equation.
[tex]Fe_2O_3+3CO\rightarrow2Fe+3CO_2[/tex]
2) Convert grams of Fe into moles of F2.
[tex]\text{mol Fe= 11.6 g Fe}_{}\cdot\frac{1\text{ mol Fe}}{55.845\text{ g Fe}}=\text{ 0.2077 mol Fe}[/tex]
0.2077 mol Fe was produced.
3) Moles of CO needed to produce 0.2077 mol Fe.
[tex]\text{mol CO}_{}=0.2077\text{ mol Fe}\cdot\frac{3\text{ mol CO}}{2\text{ mol Fe}}=0.31155\text{mol CO}[/tex]
4) Convert moles of CO into molecules of CO
[tex]\text{molecules of CO=0.31155 mol CO}\cdot\frac{6.022\cdot10^{23}}{1\text{ mol CO}}=\text{ 1.8761}\cdot10^{23}\text{ molecules of CO}[/tex]
1.88*10^23 molecules of carbon monoxide are needed to react with excess iron (III) oxide.
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