QuestionJuly 28, 2025

If 5.00 g of gray iron powder reacts with 2.38 g of red phosphorus powder, what is the mass of the iron phosphide product? square

If 5.00 g of gray iron powder reacts with 2.38 g of red phosphorus powder, what is the mass of the iron phosphide product? square
If 5.00 g of gray iron powder reacts with 2.38 g of red phosphorus powder, what is the mass of the iron phosphide
product?
square

Solution
4.0(353 votes)

Answer

4.65 g Explanation 1. Determine the limiting reactant The balanced chemical equation is 4 \text{Fe} + 3 \text{P} \rightarrow \text{Fe}_3\text{P}_2. Calculate moles of Fe: \frac{5.00 \, \text{g}}{55.85 \, \text{g/mol}} = 0.0896 \, \text{mol}. Calculate moles of P: \frac{2.38 \, \text{g}}{30.97 \, \text{g/mol}} = 0.0769 \, \text{mol}. Using stoichiometry, Fe requires \frac{3}{4} \times 0.0896 = 0.0672 \, \text{mol} of P, which is less than available, so Fe is the limiting reactant. 2. Calculate mass of iron phosphide Moles of \text{Fe}_3\text{P}_2 formed from Fe: \frac{0.0896}{4} = 0.0224 \, \text{mol}. Molar mass of \text{Fe}_3\text{P}_2 is 3(55.85) + 2(30.97) = 207.49 \, \text{g/mol}. Mass of product: 0.0224 \times 207.49 = 4.65 \, \text{g}.

Explanation

1. Determine the limiting reactant<br /> The balanced chemical equation is $4 \text{Fe} + 3 \text{P} \rightarrow \text{Fe}_3\text{P}_2$. Calculate moles of Fe: $\frac{5.00 \, \text{g}}{55.85 \, \text{g/mol}} = 0.0896 \, \text{mol}$. Calculate moles of P: $\frac{2.38 \, \text{g}}{30.97 \, \text{g/mol}} = 0.0769 \, \text{mol}$. Using stoichiometry, Fe requires $\frac{3}{4} \times 0.0896 = 0.0672 \, \text{mol}$ of P, which is less than available, so Fe is the limiting reactant.<br />2. Calculate mass of iron phosphide<br /> Moles of $\text{Fe}_3\text{P}_2$ formed from Fe: $\frac{0.0896}{4} = 0.0224 \, \text{mol}$. Molar mass of $\text{Fe}_3\text{P}_2$ is $3(55.85) + 2(30.97) = 207.49 \, \text{g/mol}$. Mass of product: $0.0224 \times 207.49 = 4.65 \, \text{g}$.
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