Question

What is the percent yield of a reaction in which 74.1 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 8.01 mL of water (d = 1.00 g/mL)?

Answer 1

Answer : The percent yield of the reaction is, 46.49 %

Explanation :  Given,

Mass of $WO_3$ = 74.1 g

Molar mass of $WO_3$ = 231.84 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $WO_3$.

$\text{Moles of }WO_3=\frac{\text{Mass of }WO_3}{\text{Molar mass of }WO_3}=\frac{74.1g}{231.84g/mole}=0.319mole$

Now we have to calculate the moles of $H_2O$.

The balanced chemical reaction will be,

$WO_3+3H_2\rightarrow W+3H_2O$

From the balanced reaction, we conclude that

As, 1 mole of $WO_3$ react to give 3 moles of $H_2O$

So, 0.319 moles of $WO_3$ react to give $3\times 0.319=0.957$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.957mole)\times (18g/mole)=17.226g$

The theoretical yield of $H_2O$  = 17.226 g

Now we have to calculate the actual yield of water.

$\text{Actual mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/ml\times 8.01ml=8.01g$

The actual yield of $H_2O$  = 8.01 g

Now we have to calculate the percent yield of $H_2O$

$\%\text{ yield of }H_2O=\frac{\text{Actual yield of }H_2O}{\text{Theoretical yield of }H_2O}\times 100=\frac{8.01g}{17.226g}\times 100=46.49\%$

Therefore, the percent yield of the reaction is, 46.49 %