Question

What is the percent yield of a reaction in which 200. g of phosphorus trichloride reacts with excess water to form 128 g of HCl and aqueous phosphorous acid$\left(H_3{\mathrm{PO}}_3\right)$ ?

Short answer

Percent yield for the given process is 80.6%.

Step-by-step solution

Writing balanced equation

First of all, let us check the balanced equation to find the percent yield:

${\mathbf{PCI}}_{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\stackrel{\mathbf{}}{\mathbf{\to }}{\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{3}}\mathbf{+}\mathbf{3}\mathbf{HCI}$

From the equation we can say that 1 mol of PCl₃ gives three moles of HCl.

Calculating Moles of PCl3

Moles can be calculated as

$\mathbf{Moles}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{PCI}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{PCI}}_{\mathbf{3}}}{\mathbf{Molar}\mathbf{}\mathbf{mass}}\mathbf{=}\frac{\mathbf{200}}{\mathbf{137}\mathbf{.}\mathbf{33}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}$

From the balanced equation we can say that,

$\mathbf{3}\mathbf{\times }\mathbf{Moles}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{PCI}}_{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{Moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{HCI}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{35}$

Theoretical yield

The maximum possible mass of a product that can be formed in a chemical reaction, is known as its theoretical yield. Hence, Theoretical yield of HCl is

$\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{HCI}\mathbf{=}\mathbf{Moles}\mathbf{\times }\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{35}\mathbf{\times }\mathbf{36}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{158}\mathbf{.}\mathbf{775}\mathbf{}\mathbf{g}$

Percent yield calculation

Mass of HCl actually formed = 128 g (given)

Percent yield can be calculated using the given formula:

$\mathbf{Percent}\mathbf{}\mathbf{yield}\mathbf{=}\frac{\mathbf{Actual}\mathbf{}\mathbf{yield}}{\mathbf{theoritical}\mathbf{}\mathbf{yield}}\mathbf{\times }\mathbf{100}\mathbf{=}\frac{\mathbf{128}}{\mathbf{158}\mathbf{.}\mathbf{775}}\mathbf{\times }\mathbf{100}\mathbf{=}\mathbf{80}\mathbf{.}\mathbf{6}\mathbf{\%}$

So, Percent yield for the given process is 80.6%.