Question

If 9.95 g of carbon dioxide is produced from the reaction of 6.90 g of methane and 20.1 g of oxygen gas, calculate the percent yield of carbon dioxide.

Answer 1

CH4(g) + 2O2(g) -------------> CO2(g) + 2H2O(g)

no of moles of CH4 = W/G.M.Wt

                                = 6.9/16   = 0.43125 moles

no of mooles of O2   = W/G.M.Wt

                               = 20.1/32   = 0.628125moles

CH4(g) + 2O2(g) -------------> CO2(g) + 2H2O(g)

1 mole of CH4 react with 2 moles of O2

0.43125 moles of CH4 react with = 2*0.43125/1   = 0.8625 moles of O2 is required

O2 is limiting reactant

2 moles of O2 react with excess of CH4 to gives 1 mole of CO2

0.628125 moles of O2 react with excess of CH4 to gives = 1*0.628125/2   = 0.314 moles of CO2

mass of CO2   = no of moles * gram molar mass

                        = 0.314*44   = 13.816g

Theoretical yield   = 13.816g

actual yield           = 9.95g

percentage yield of CO2     = actual yield *100/theoretical yield

                                             = 9.95*100/13.816   = 72% >>>>>answer