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 89%. suppose 5.0 g of a are reacted with excess compound b, and 8.5 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: Calculate theoretical yield of \( C \)

The formula for percent yield is \( \text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\% \).
We know that percent yield \( = 89\%=0.89 \), actual yield \( = 8.5\ \text{g} \).
Let the theoretical yield be \( x \). Then \( 0.89=\frac{8.5}{x} \), so \( x=\frac{8.5}{0.89}\approx9.6\ \text{g} \).

Step2: Calculate mass of \( B \) consumed

Since \( A + B
ightarrow C \), from the law of conservation of mass (in a reaction where \( A \) reacts with \( B \) to form \( C \) and \( B \) is in excess).
The mass of \( A\) is \( 5.0\ \text{g} \), the theoretical mass of \( C \) is approximately \( 9.6\ \text{g} \).
By the law of conservation of mass \( m(A)+m(B)=m(C) \) (theoretical).
So \( m(B)=m(C)-m(A) \).
\( m(B)=9.6 - 5.0=4.6\ \text{g} \)

Answer:

• Theoretical yield of \( C \): \( 9.6\ \text{g} \)
• Mass of \( B \) consumed: \( 4.6\ \text{g} \)