QuestionApril 24, 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
4.6(243 votes)

Answer

5.56 g Explanation 1. Determine the limiting reactant Calculate moles of iron (Fe) and phosphorus (P). Molar mass of Fe = 55.85 \, g/mol, P = 30.97 \, g/mol. Moles of Fe = \frac{5.00}{55.85} \approx 0.0895 \, mol. Moles of P = \frac{2.38}{30.97} \approx 0.0769 \, mol. The reaction is 3Fe + 2P \rightarrow Fe_3P. Compare mole ratio: \frac{0.0895}{3} \approx 0.0298 for Fe, \frac{0.0769}{2} \approx 0.0385 for P. Iron is the limiting reactant. 2. Calculate mass of iron phosphide (Fe_3P) Use limiting reactant moles to find product mass. Moles of Fe_3P = 0.0298 \, mol. Molar mass of Fe_3P = 3(55.85) + 30.97 = 186.52 \, g/mol. Mass of Fe_3P = 0.0298 \times 186.52 \approx 5.56 \, g.

Explanation

1. Determine the limiting reactant<br /> Calculate moles of iron ($Fe$) and phosphorus ($P$). Molar mass of $Fe = 55.85 \, g/mol$, $P = 30.97 \, g/mol$. Moles of $Fe = \frac{5.00}{55.85} \approx 0.0895 \, mol$. Moles of $P = \frac{2.38}{30.97} \approx 0.0769 \, mol$. The reaction is $3Fe + 2P \rightarrow Fe_3P$. Compare mole ratio: $\frac{0.0895}{3} \approx 0.0298$ for $Fe$, $\frac{0.0769}{2} \approx 0.0385$ for $P$. Iron is the limiting reactant.<br />2. Calculate mass of iron phosphide ($Fe_3P$)<br /> Use limiting reactant moles to find product mass. Moles of $Fe_3P = 0.0298 \, mol$. Molar mass of $Fe_3P = 3(55.85) + 30.97 = 186.52 \, g/mol$. Mass of $Fe_3P = 0.0298 \times 186.52 \approx 5.56 \, g$.
Click to rate:

Similar Questions