Question

in a certain chemical reaction compound a combines with compound b to produce compound c (and no other products). measurements were taken of the amounts of a and b present before and after a reaction that produced some c:

calculate the theoretical yield of c.
round your answer to the nearest 0.1 g.

suppose the percent yield of c in this reaction was 76.%. calculate the actual amount of c that was isolated at the end of the reaction.
round your answer to the nearest 0.1 g.

Explanation:

Step1: Calculate the mass of reactants consumed

• Mass of \(A\) consumed: \(m_A=9.0 - 0=9.0\) g
• Mass of \(B\) consumed: \(m_B = 6.0 - 4.4=1.6\) g

Step2: Calculate theoretical yield of \(C\)

According to the law of conservation of mass (in a chemical reaction, mass is conserved, \(m_{reactants}=m_{products}\)), the theoretical yield of \(C\), \(m_{C,theo}\) is the sum of the masses of \(A\) and \(B\) that reacted.
\(m_{C,theo}=m_A + m_B\)
\(m_{C,theo}=9.0+1.6 = 10.6\) g

Step3: Calculate actual yield of \(C\)

The percent - yield formula is \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\)
Let the actual yield be \(m_{C,act}\). We know that \(\text{Percent Yield} = 76\%\) and \(m_{C,theo}=10.6\) g
\(76\%=\frac{m_{C,act}}{10.6}\times 100\%\)
\(m_{C,act}=\frac{76\times10.6}{100}\)
\(m_{C,act}=\frac{805.6}{100}=8.056\approx8.0\) g (rounded to the nearest \(0.1\) g)

Answer:

For theoretical yield: \(10.6\) g
For actual yield: \(8.0\) g