Given 10.0 g of PCl3 and 12.0 g of Cl2, what mass of PCl5 can be produced according to the reaction PCl3 + Cl2 → PCl5, assuming the reaction goes to completion?

• A. 10.0 g
• B. 22.2 g
• C. 17.2 g
• D. 12.0 g

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction: \(\mathrm{PCl_3 + Cl_2 \rightarrow PCl_5}\).
2. Calculate the molar masses of each substance: \(\mathrm{PCl_3}\), \(\mathrm{Cl_2}\), and \(\mathrm{PCl_5}\). Use atomic masses (P = 30.97 g/mol, Cl = 35.45 g/mol) to find these values.
3. Convert the given masses of \(\mathrm{PCl_3}\) (10.0 g) and \(\mathrm{Cl_2}\) (12.0 g) to moles using their molar masses: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
4. Determine the limiting reactant by comparing the mole ratio of the reactants to the stoichiometric ratio from the balanced equation (1:1). The reactant that produces fewer moles of \(\mathrm{PCl_5}\) is the limiting reactant.
5. Calculate the mass of \(\mathrm{PCl_5}\) produced by multiplying the moles of the limiting reactant by the molar mass of \(\mathrm{PCl_5}\): \(\text{mass} = \text{moles} \times \text{molar mass}\).