Answer:
The percent yield of reaction is 54.32%.
Explanation:
\(CH_4+3O_2\rightarrow CO_2+2H_2O\)
Moles of methane = \(\frac{5.00 g}{16.05 g/mol}=0.3115 mol\)
According to reaction, 1 mole of methane gives 2 moles of water .
The 0.3115 moles of methane will give:
\(\frac{2}{1}\times 0.3115 mol=0.623 mol\) of water
Mass of 0.9345 moles of water = 0.623 mol × 18.02 g/mol = 11.23 g.
Theoretical yield of methane = 11.23 g.
Experimental yield of methane = 6.10 g.
The percent yield of reaction:
\(\%(Yield)=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100\)
The percent yield of reaction is :
\(\%(Yield)=\frac{6.10 g }{11.23 g}\times 100=54.32\%\)
