Q84P - Chemistry: Molecular Nature Of Matter And Change

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  $\left(H_3{\mathrm{PO}}_3\right)$ ?

Step-by-Step Solution

Answer

Percent yield for the given process is 80.6%.

Step 1: 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.

Step 2: Calculating Moles of PCl 3

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,

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

Step 3: 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 

 

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

Step 4: 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%.