Answer :
We begin by determining how many moles of benzene ([tex]\( C_6H_6 \)[/tex]) are present in [tex]\( 36.8\text{ g} \)[/tex]. The molar mass of benzene is approximately
[tex]$$
M_{C_6H_6} \approx 78.0\text{ g/mol}.
$$[/tex]
Thus, the number of moles of benzene is calculated by
[tex]$$
n_{C_6H_6} = \frac{36.8\text{ g}}{78.0\text{ g/mol}} \approx 0.4718\text{ mol}.
$$[/tex]
Since the balanced chemical equation
[tex]$$
C_6H_6(l) + Cl_2(g) \rightarrow C_6H_5Cl(l) + HCl(g)
$$[/tex]
shows a 1:1 molar ratio between benzene and chlorobenzene ([tex]\( C_6H_5Cl \)[/tex]), there will be the same number of moles of [tex]\( C_6H_5Cl} \)[/tex] formed theoretically.
Next, we calculate the theoretical yield of [tex]\( C_6H_5Cl \)[/tex]. The molar mass of chlorobenzene is approximately
[tex]$$
M_{C_6H_5Cl} \approx 112.5\text{ g/mol}.
$$[/tex]
Thus, the theoretical mass of [tex]\( C_6H_5Cl \)[/tex] produced is
[tex]$$
\text{Theoretical yield} = n_{C_6H_5Cl} \times M_{C_6H_5Cl} = 0.4718\text{ mol} \times 112.5\text{ g/mol} \approx 53.08\text{ g}.
$$[/tex]
Given that the actual yield is [tex]\( 38.8\text{ g} \)[/tex], the percentage yield is calculated as follows:
[tex]$$
\text{Percentage yield} = \left(\frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% = \left(\frac{38.8\text{ g}}{53.08\text{ g}}\right) \times 100\% \approx 73.10\%.
$$[/tex]
Thus, the percentage yield of [tex]\( C_6H_5Cl \)[/tex] is approximately [tex]\( 73.1\% \)[/tex].
[tex]$$
M_{C_6H_6} \approx 78.0\text{ g/mol}.
$$[/tex]
Thus, the number of moles of benzene is calculated by
[tex]$$
n_{C_6H_6} = \frac{36.8\text{ g}}{78.0\text{ g/mol}} \approx 0.4718\text{ mol}.
$$[/tex]
Since the balanced chemical equation
[tex]$$
C_6H_6(l) + Cl_2(g) \rightarrow C_6H_5Cl(l) + HCl(g)
$$[/tex]
shows a 1:1 molar ratio between benzene and chlorobenzene ([tex]\( C_6H_5Cl \)[/tex]), there will be the same number of moles of [tex]\( C_6H_5Cl} \)[/tex] formed theoretically.
Next, we calculate the theoretical yield of [tex]\( C_6H_5Cl \)[/tex]. The molar mass of chlorobenzene is approximately
[tex]$$
M_{C_6H_5Cl} \approx 112.5\text{ g/mol}.
$$[/tex]
Thus, the theoretical mass of [tex]\( C_6H_5Cl \)[/tex] produced is
[tex]$$
\text{Theoretical yield} = n_{C_6H_5Cl} \times M_{C_6H_5Cl} = 0.4718\text{ mol} \times 112.5\text{ g/mol} \approx 53.08\text{ g}.
$$[/tex]
Given that the actual yield is [tex]\( 38.8\text{ g} \)[/tex], the percentage yield is calculated as follows:
[tex]$$
\text{Percentage yield} = \left(\frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% = \left(\frac{38.8\text{ g}}{53.08\text{ g}}\right) \times 100\% \approx 73.10\%.
$$[/tex]
Thus, the percentage yield of [tex]\( C_6H_5Cl \)[/tex] is approximately [tex]\( 73.1\% \)[/tex].
