QuestionJune 17, 2025

PCl_(5) can be produced by the reaction PCl_(3)+Cl_(2)arrow PCl_(5) What mass of PCl_(3) must be used to produce 479.525 g of PCl_(5) if the percent yield is 62.9% square

PCl_(5) can be produced by the reaction PCl_(3)+Cl_(2)arrow PCl_(5) What mass of PCl_(3) must be used to produce 479.525 g of PCl_(5) if the percent yield is 62.9% square
PCl_(5) can be produced by the reaction PCl_(3)+Cl_(2)arrow PCl_(5) What
mass of PCl_(3) must be used to produce 479.525 g of PCl_(5) if the
percent yield is 62.9% 
square

Solution
4.5(306 votes)

Answer

503.0 g of PCl_3 must be used. Explanation 1. Calculate the theoretical yield of PCl_5 Use the molar mass of PCl_5 (208.24 g/mol) to find moles: \text{moles of } PCl_5 = \frac{479.525 \, \text{g}}{208.24 \, \text{g/mol}}. 2. Adjust for percent yield Actual moles produced = Theoretical moles \times \frac{62.9}{100}. 3. Calculate moles of PCl_3 needed From the balanced equation, 1 mole of PCl_3 produces 1 mole of PCl_5. Therefore, moles of PCl_3 needed = Actual moles of PCl_5. 4. Convert moles of PCl_3 to mass Use the molar mass of PCl_3 (137.32 g/mol): Mass of PCl_3 = \text{moles of } PCl_3 \times 137.32 \, \text{g/mol}.

Explanation

1. Calculate the theoretical yield of $PCl_5$<br /> Use the molar mass of $PCl_5$ (208.24 g/mol) to find moles: $\text{moles of } PCl_5 = \frac{479.525 \, \text{g}}{208.24 \, \text{g/mol}}$.<br />2. Adjust for percent yield<br /> Actual moles produced = Theoretical moles $\times \frac{62.9}{100}$.<br />3. Calculate moles of $PCl_3$ needed<br /> From the balanced equation, 1 mole of $PCl_3$ produces 1 mole of $PCl_5$. Therefore, moles of $PCl_3$ needed = Actual moles of $PCl_5$.<br />4. Convert moles of $PCl_3$ to mass<br /> Use the molar mass of $PCl_3$ (137.32 g/mol): Mass of $PCl_3 = \text{moles of } PCl_3 \times 137.32 \, \text{g/mol}$.
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