Iron reacts with oxygen to produce iron (III) oxide. 4 Fe (s) + 3 O_2 (g) to 2 Fe_2O_3 (s) If 6.4 moles of Fe react with excess O_2, how many moles of Fe_2O_3 can be formed?

The balanced chemical equation for the reaction between iron and oxygen is: {eq}4 \text{ Fe} (s) + 3 \text{ O}_2 (g) rightarrow 2 \text{ Fe}_2\text{ O}_3 (s){/eq} This equation tells us that four moles of iron react with three moles of oxygen to produce two moles of iron (III) oxide. If 6.4 moles of iron are reacted with excess oxygen, we can use stoichiometry to determine how many moles of iron (III) oxide are produced. We can set up a proportion based on the coefficients in the balanced equation: {eq}frac{4 \text{ mol Fe}}{2 \text{ mol Fe}_2\text{ O}_3} = \frac{6.4 \text{ mol Fe}}{x \text{ mol Fe}_2\text{ O}_3}{/eq} Solving for x, the number of moles of iron (III) oxide produced: {eq}x = \frac{2 \text{ mol Fe}_2\text{ O}_3 times 6.4 \text{ mol Fe}}{4 \text{ mol Fe}} = 3.2 \text{ mol Fe}_2\text{ O}_3{/eq} Therefore, if 6.4 moles of iron react with excess oxygen, 3.2 moles of iron (III) oxide would be produced.