Stoichiometry Question

Question

During the combustion of methane, what mass of carbon dioxide will be produced when 15.0g of CH4(g) and 30.0(g) of O2(g) are available for reaction? If 17.55g of carbon dioxide was obtained during the experiment, what was the percent yield of the reaction?

Answer 1

Combustion of a hydrocarbon, like methane,produces carbon dioxide and water. The balanced equation is: CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g). The molecular weight of methane is 12 + 4 = 16. So 15.0 grams of methane represents 15.0/16.0 moles = 0.9375 moles. And 30.0 grams of oxygen represents 30.0grams/32.0 grams/moles = 0.9375 moles. From the balanced equation we see that for every mole of methane combusted, we need 2 moles of oxygen. Because we have equal number of moles of methane and oxygen, only half of the methane can be combusted in the manner shown by the equation. Accordingly, assuming all the oxygen is used up in the combustion of the methane, only 0.9375/2 moles of carbon dioxide will be produced = 0.4687 moles of CO₂. Since the molecular weight of CO₂ is 12 + 32 = 44, the mass of CO₂ produced under these conditions = 0.4687 x 44 = 20.62 grams. Collection of only 17.55 grams of CO₂

means a yield of 17.55/20.62 = 85.11% yield.