Final answer:
To calculate the percent yield, we determined the limiting reactant using stoichiometry, found the theoretical yield of CO₂, and then compared it to the actual yield. The calculation resulted in a percent yield of approximately 90.3%, which aligns most closely with option d (91.2%).
Step-by-step explanation:
To find the percent yield of the reaction 2CO(g) + O₂(g) ⇒ 2CO₂(g), we compare the actual yield to the theoretical yield. First, we need to determine the limiting reactant, which is the reactant that will be completely consumed and thus limits the amount of products formed. We can use stoichiometry based on the balanced chemical equation.
The molar mass of CO is 28.01 g/mol and for O₂ it is 32.00 g/mol. Thus, 21.0 g CO is equivalent to 21.0 g / 28.01 g/mol = 0.75 mol of CO, and 12.8 g O₂ is equivalent to 12.8 g / 32.00 g/mol = 0.40 mol of O₂. According to the reaction stoichiometry, 1 mol of O2 reacts with 2 mol of CO, hence O2 is the limiting reactant as we have less than the required 0.75 mol (as only 0.40 mol is available).
The theoretical yield is determined by the amount of limiting reactant, in this case, O₂, which can form 0.40 mol x 2 = 0.80 mol CO₂. Converting moles of CO₂ to grams, using its molar mass (44.01 g/mol), we get 0.80 mol x 44.01 g/mol = 35.21 g of CO₂ as the theoretical yield. The actual yield given is 31.8 g CO₂.
To calculate the percent yield, we use the formula Percent yield = (Actual Yield / Theoretical Yield) x 100. Substituting the values, we get Percent Yield = (31.8 g / 35.21 g) x 100 = 90.3%, which is not exactly one of the provided options but is closest to option d.