Respuesta :
Answer : The percent yield of the reaction is, 46.49 %
Explanation : Given,
Mass of [tex]WO_3[/tex] = 74.1 g
Molar mass of [tex]WO_3[/tex] = 231.84 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]WO_3[/tex].
[tex]\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[/tex]
Now we have to calculate the moles of [tex]H_2O[/tex].
The balanced chemical reaction will be,
[tex]WO_3+3H_2\rightarrow W+3H_2O[/tex]
From the balanced reaction, we conclude that
As, 1 mole of [tex]WO_3[/tex] react to give 3 moles of [tex]H_2O[/tex]
So, 0.319 moles of [tex]WO_3[/tex] react to give [tex]3\times 0.319=0.957[/tex] moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex]
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(0.957mole)\times (18g/mole)=17.226g[/tex]
The theoretical yield of [tex]H_2O[/tex] = 17.226 g
Now we have to calculate the actual yield of water.
[tex]\text{Actual mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/ml\times 8.01ml=8.01g[/tex]
The actual yield of [tex]H_2O[/tex] = 8.01 g
Now we have to calculate the percent yield of [tex]H_2O[/tex]
[tex]\%\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\%[/tex]
Therefore, the percent yield of the reaction is, 46.49 %
