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)?

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.

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,

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

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:

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

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) = $\mathbf{0}\mathbf{.}\mathbf{178}\mathbf{\times }\mathbf{184}\mathbf{=}\mathbf{32}\mathbf{.}\mathbf{752}\mathbf{}\mathbf{g}$

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%.