\(2 C _{4} H _{10}+13 O _{2} \rightarrow 8 CO _{2}+10 H _{2} O\)
M.M of butane \(\Rightarrow(12 \times 4)+(1 \times 10) \Longrightarrow 58 gmol ^{-1}\)
Mole of butane \(\Rightarrow\left(\frac{299}{58 \text { gmol }^{-1}}\right)\)
\(\Rightarrow 0.5 mol\)rom the equation of reaction,
2 moles of \(C _{4} H _{10}\) produces 10 moles of \(H _{2} O\)
\(0.5\) mole of \(C _{4} H _{10}\) produces
\(\left(\frac{299}{2} \times 0.5\right)\) mole \(=2.5\) moles
Mass of water produced theoretically
\(=(2.5 \times 18) \Rightarrow 45 g\) Actual yield of \(H _{2} O \Rightarrow 0.90 g\)
\(\%\) yield \(=\frac{\text { Actual yield }}{\text { Theoretical }} \times 100 \%\)
\(\Rightarrow \frac{0.90 g }{45 g } \times 100 \%\)
\(\Rightarrow 2 \%\)
You can also use the ratio of moles
\(0.90 g (0.90 / 18)\) mole \(\Longrightarrow 0.05\) mole
of \(H _{2} O\) of \(H _{2} O\)
\(\%\) yield \(=\frac{0.05}{2.5} \times 100 \% \Rightarrow 2 \%\)