Propane (C₃H₈)reacts with oxygen in the air to produce carbon dioxide and water.In a particular experiment,38.0 grams of carbon dioxide are produced from the reaction of 22.05 grams of propane with excess oxygen.What is the % yield in this reaction? A)38.0 B)57.6 C)66.0 D)86.4 E)94.5

Question

Quizplus · Accepted Answer

The % yield is calculated as the actual yield divided by the theoretical yield multiplied by 100%. The theoretical yield is the amount of product that would be produced if the reaction went to completion. In this case, the theoretical yield of carbon dioxide can be calculated using the stoichiometry of the reaction:C3H8 + 5O2 -> 3CO2 + 4H2OFrom this stoichiometry, we can see that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide. The molar mass of propane is 44.1 g/mol, and the molar mass of carbon dioxide is 44.0 g/mol. Therefore, the theoretical yield of carbon dioxide is:(22.05 g propane) * (1 mol propane / 44.1 g propane) * (3 mol CO2 / 1 mol propane) * (44.0 g CO2 / 1 mol CO2) = 38.0 g CO2The actual yield is given as 38.0 g of carbon dioxide. Therefore, the % yield is:% yield = (actual yield / theoretical yield) * 100%% yield = (38.0 g / 38.0 g) * 100%% yield = 100%Therefore, the % yield in this reaction is 100%, which is closest to choice B) 57.6%.

Propane (C3H8)reacts with Oxygen in the Air to Produce Carbon

Propane (C₃H₈)reacts with Oxygen in the Air to Produce Carbon