Q83P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the percent yield of a reaction in which 45.5 g of tungsten (VI) oxide (WO₃) reacts with excess hydrogen gas to produce metallic tungsten and 9.60 mL of water (d = 1.00 g/mL)?

Step-by-Step Solution

Answer

Percent yield for the given process is 91%.

Step 1: Writing balanced equation

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

 

${\mathrm{WO}}_3+3{\mathrm{H}}_2 \stackrel{ }{\to } \mathrm{W}+3{\mathrm{H}}_2\mathrm{O}$ 

From the equation we can say that 1 mol of  ${\mathbf{WO}}_{\mathbf{3}}$ gives one mole of W.

Step 2: Calculating Moles of WO 3

Moles can be calculated as 

 

$\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{WO}}_{\mathbf{3}}\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{WO}}_{\mathbf{3}}}{\mathbf{Molar}\mathbf{ }\mathbf{mass}}\mathbf{=}\frac{\mathbf{45}\mathbf{.}\mathbf{5}}{\mathbf{232}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{196}$ 

 

From the balanced equation we can say that, 

 

 $\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{WO}}_{\mathbf{3}}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{W}\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{196}$

Step 3: Determine the 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 tungsten is:

 

 $\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{W}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{Mole}\mathbf{\times }\mathbf{Molar}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{196}\mathbf{\times }\mathbf{184}\mathbf{=}\mathbf{36}\mathbf{ }\mathbf{g}$

Step 4: Actual yield Calculation

As density of water = 1 g/mL so mass of water =9.6 g

 

$\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}}\mathbf{=}\frac{\mathbf{9}\mathbf{.}\mathbf{6}}{\mathbf{18}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{53}$ 

 

From the equation we can say that 1 mole tungsten (184 g) is formed when 3 moles of water are formed.

So if 0.53 moles of water are formed then moles of tungsten formed are: 0.178 mole

Mass of tungsten actually formed (actual yield)  =     $\mathbf{0}\mathbf{.}\mathbf{178}\mathbf{\times }\mathbf{184}\mathbf{=}\mathbf{32}\mathbf{.}\mathbf{752}\mathbf{ }\mathbf{g}$

Step 5: Percent yield calculation

Percent yield can be calculated using the given formula:

 

$\mathbf{Percent}\mathbf{ }\mathbf{yield}\mathbf{=}\frac{\mathbf{Actual}\mathbf{ }\mathbf{yield}}{\mathbf{theoritical}\mathbf{ }\mathbf{yield}}\mathbf{\times }\mathbf{100}\mathbf{=}\frac{\mathbf{32}\mathbf{.}\mathbf{752}}{\mathbf{36}}\mathbf{\times }\mathbf{100}\mathbf{=}\mathbf{91}\mathbf{\%}$ 

 

So, Percent yield for the given process is 91%.