Question

Methane (CH4, 16.05 g/mol) reacts with oxygen to form carbon dioxide (CO2, 44.01 g/mol) and water (H2O, 18.02 g/mol). Assume that you design a system for converting methane to carbon dioxide and water. To test the efficiency of the system in the laboratory, you burn 5.00 g methane. The actual yield is 6.10 g water. What is your percent yield?

Answer 1

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 :

$\%(Yield)=\frac{6.10 g }{11.23 g}\times 100=54.32\%$

Answer 2

Answer:54.5

Explanation: