If 42.0 g of HCl are produced from the reaction of 62.0 g of PCl_3, what is the percent yield of HCl? (Assume the reaction is: PCl_3 + 3 H_2O → H_3PO_3 + 3 HCl. Molar masses: PCl_3 = 137.33 g/mol, HCl = 36.46 g/mol.)

• A. 67.2%
• B. 100%
• C. 81.5%
• D. 45.2%

Answer: (A)

Verified step by step guidance

1. Write the balanced chemical equation: \(\text{PCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 3 \text{HCl}\).
2. Calculate the moles of PCl\(_3\) reacted using its molar mass: \(\text{moles of PCl}_3 = \frac{62.0\ \text{g}}{137.33\ \text{g/mol}}\).
3. Use the stoichiometric ratio from the balanced equation to find the theoretical moles of HCl produced: since 1 mole of PCl\(_3\) produces 3 moles of HCl, multiply the moles of PCl\(_3\) by 3.
4. Calculate the theoretical mass of HCl produced by multiplying the theoretical moles of HCl by its molar mass: \(\text{mass of HCl}_{\text{theoretical}} = \text{moles of HCl} \times 36.46\ \text{g/mol}\).
5. Calculate the percent yield using the formula: \(\text{percent yield} = \left( \frac{\text{actual mass of HCl}}{\text{theoretical mass of HCl}} \right) \times 100\%\), where the actual mass of HCl is given as 42.0 g.