Question

compound a reacts with compound b to form only one product, compound c, and its known the usual percent yield of c in this reaction is 87.%. suppose 4.5 g of a are reacted with excess compound b, and 5.7 g of compound c are successfully isolated at the end of the reaction. what was the theoretical yield of c? round your answer to the nearest 0.1 g. how much b was consumed by the reaction? round your answer to the nearest 0.1 g.

Explanation:

Step1: Recall percent yield formula

The percent yield formula is $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$. We know the percent yield (87.%) and the actual yield (5.7 g), so we can solve for theoretical yield. Rearranging the formula: $\text{Theoretical Yield} = \frac{\text{Actual Yield}}{\text{Percent Yield} / 100\%}$.

Step2: Calculate theoretical yield of C

Substitute the values: $\text{Theoretical Yield} = \frac{5.7\ \text{g}}{0.87}$. Calculate this: $5.7 \div 0.87 \approx 6.6\ \text{g}$ (rounded to nearest 0.1 g).

Step3: Use law of conservation of mass for B

In a reaction with excess B, the mass of A plus mass of B consumed equals mass of C (theoretical yield). So mass of B consumed = theoretical yield of C - mass of A. Mass of A is 4.5 g, theoretical yield of C is ~6.6 g. So $6.6 - 4.5 = 2.1\ \text{g}$.

Answer:

Theoretical yield of C: $\boxed{6.6}$ g
Mass of B consumed: $\boxed{2.1}$ g