Question

1) phosphorus pentachloride is produced by the reaction between phosphine and chlorine gas. What mass of chlorine gas would be required to produce 0.015kg of phosphorus pentachloride
2) if 10g of phosphine is made to react with 15g of chlorine gas. Calculate the percentage yield if 5.5g of phosphorus pentachloride was experimentally produced

Answer 1

1)

$$4Cl_2 + PH_3 \rightarrow PCl_5 + 3HCl$$

$n = \frac{m}{M}$

$n(PCl_5) = \frac{15 g}{208.22 \frac{g}{mo}} = 0.072 mol$

According to equation mole ratio $n(Cl_2):n(PCl_5) = 4:1$, then $n(Cl_2) = n(PCl_5) \times 4 = 0.072 mol \times 4 = 0.288 mol$

$m(Cl_2) = n(Cl_2)\times M(Cl_2) = 0.288 mol \times 70.9 \frac{g}{mol} = 20.4 g = 0.0204 kg$

2)

$$4Cl_2 + PH_3 \rightarrow PCl_5 + 3HCl$$

$n(Cl_2) = \frac{m}{M} = \frac {15 g}{70.9 \frac{g}{mol}}= 0.212 mol$

$n(PH_3) = \frac{m}{M} = \frac{10 g}{34\frac{g}{mo}} = 0.294 mol$

Compare ratios $\frac{n(Cl_2)}{4} = \frac{0.212 mol}{4} = 0.053 mol$ and $\frac{n(PH_3)}{1} = 0.294 mol$

$0.053 mol < 0.294 mol$ , then $Cl_2$ is a limiting rectant.

Use $n(Cl_2) = 0.212 mol$ to determine $n(PCl_5)$

According to equation mole ratio $n(Cl_2): n(PCl_5) = 4:1$ , then $n(PCl_5) = \frac{n(Cl_2)}{4} = \frac{0.212 mol}{4} = 0.053 mol$

$m_{theor}(PCl_5) = n\times M = 0.053 mol \times 208.22 \frac{g}{mol} = 11.04 g$

percentage yield % $= \frac{5.5 g}{11.04 g} \times 100\% = 49.8 \%$