Question

When 60.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 112 g. What is the percent yield? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Group of answer choices

Answer 1

Answer:

67.88% is the percent yield.

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g)$

Moles of methane = $\frac{60.0 g}{16 g/mol}=3.75 mol$

According to reaction,1 mole of methane gives 1 mole of carbon dioxide gas, then 3.75 moles of methane will give :

$\frac{1}{1}\times 3.75 mol =3.75 mol$ of carbon dioxide gas

Mass of 3.75 moles of carbon dioxide gas:

3.75 mol × 44 g/mol = 165 g

Theoretical yield of the carbon dioxide gas = 165 g

Experimental yield of the carbon dioxide gas = 112 g

The percentage yield of the reaction :

$Yield\%=\frac{\text{actual yield}}{\text{Theoretical yield}}\times 100$

$=\frac{112 g}{165 g}\times 100=67.88\%$

67.88% is the percent yield.