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?

Explanation:

Step1: Write the balanced chemical equation

$$\ce{CH4 + 2O2->CO2 + 2H2O}$$

Step2: Calculate the moles of methane

Moles of \(CH_4=\frac{mass}{molar\ mass}=\frac{5.00\ g}{16.05\ g/mol}\approx0.3115\ mol\)

Step3: Calculate the theoretical moles of water

From the balanced equation, mole ratio of \(CH_4:H_2O = 1:2\). So moles of \(H_2O = 2\times0.3115\ mol = 0.623\ mol\)

Step4: Calculate the theoretical mass of water

Mass of \(H_2O=moles\times molar\ mass = 0.623\ mol\times18.02\ g/mol\approx11.23\ g\)

Step5: Calculate the percent yield

Percent yield\(=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{6.10\ g}{11.23\ g}\times100\%\approx54.3\%\)

Answer:

\(54.3\%\)