Answer to Question #291111 in General Chemistry for hargun

Question

62. Methylbenzene (toluene), CHI;CH;(1), is a common

paint thinner. It can be prepared from benzene, Cattal

CH,CI(g) + CoHo(1) > CH;CH;() + HCK(g)

In an investigation, 25.0 g of benzene is combined

with 20.0 g of chloromethane, CH, C1, in the presence

of an appropriate catalyst. (7.4, 7.5)

(a) Calculate the theoretical yield of toluene.

(b) If 22.0 g of CH, CH, is actually obtained, what is the

percentage yield?

(c)

Suggest reasons for the discrepancy between the

actual yield and the theoretical yield.

Answer 1

$CH_3Cl_{(g)}+C_6H_{6(l)}\to\>C_6H_5CH_{3(l)}+HCl_{(g)}$

Molar mass of CH$_3$ Cl $=50.49g/mol$

C₆H₆ $=78.11g/mol$

C₆H₅CH₃ $=92.14g/mol$

Moles of benzene $=\frac{25}{78.11}=0.3201$

Moles of CH₃Cl$=\frac{20}{50.49}=0.3961$

Benzene is the limiting reactant

Moles of Toluene $=$ moles of Benzene

$=0.3201$

Theoretical yield $=0.3201×92.14$

$=29.49059$

(b)

Percentage yield $=\frac{22}{29.4905}×100\%$

$=74.6\%$

(c)

Some of the materials;

(1) Were wasted

(2) They failed to react

(3) The product were not captured