In an experiment, 8.50 g of methane, CH4, was reacted with 15.9 g of oxygen gas to produce carbon dioxide and water. Determine the percentage yield if 9.77 g of carbon dioxide was obtained in the lab.

89.3 %

Explanation:

M(CH4) = 12+ 4*1 = 16 g/mol

M(O2) = 2*16 = 32 g/mol

M(CO2) = 12 + 2*16 = 44 g/mol

8.50 g * 1 mol/16 g = 0.5313 mol CH4

15.9 g * 1 mol/32 g = 0.4969 mol O2

9.77 g * 1 mol/44 g = 0.2220 mol CO2

1)                                  CH4        + 2O2 > CO2 + 2H2O

from reaction             1 mol          2 mol

given                       0.5313 mol   (0.4969 mol)

1 mol CH4              --- 2 mol O2

0.5313 mol  CH4  --- x mol O2

x= 2*0.5313 = 1.0626 mol O2

We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.

2)                               CH4        + 2O2 > CO2 + 2H2O

from reaction                             2 mol        1 mol

given                                      0.4969 mol   x mol

x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield

3)

Practical yield CO2 = 0.2220 mol

Theoretical yield CO2 = 0.2485 mol

% yield = (0.2220/0.2485)*100% = 89.3 %

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