Answer : The percent yield of \(H_2O\) is, 45.97 %
Explanation :
First we have to calculate the moles of water.
The balanced chemical reaction is,
\(2H_2+O_2\rightarrow 2H_2O\)
From the balanced reaction, we conclude that
As, 1 moles of \(O_2\) react to give 2 moles of \(H_2O\)
So, 2 moles of \(O_2\) react to give \(2\times 2=4\) moles of \(H_2O\)
Now we have to calculate the mass of \(H_2O\).
\(\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O\)
\(\text{Mass of }H_2O=(4mole)\times (18g/mole)=72g\)
The mass water produces, 72 g
Now we have to calculate the percent yield of \(H_2O\).
\(\%\text{ yield of }H_2O=\frac{\text{Actual yield of }H_2O}{\text{Theoretical yield of }H_2O}\times 100=\frac{33.1g}{72g}\times 100=45.97\%\)
Therefore, the percent yield of \(H_2O\) is, 45.97 %
