In the reaction

C_3H_7Br + KOH → C_3H_7OH + KBr

24.6 g of 1 bromopropane reacts with excess potassium hydroxide to produce 8.00 g of propan-1-ol.

M_r : C3H7Br = 123

A_r : H = 1.0; C = 12.0; O = 16.0

What is the percentage yield of this reaction?

A. 57.1%

B. 93.0%

C. 32.5%

D. 33.3%

E. 66.7%

1 Like

AhdOmer
#2
This reaction is simpler than it looks, and it begins with the fact that the molar ratio for the reactants to products is 1:1.

When calculating percentage yield, we must find out the actual amount of product produced, and the theoretical amount according to the formula given.

We know that the ratio of 1 bromopropane and propan-1-ol is 1:1.

The actual amount produced of 1 bromopropane is 24.6 g.

In theory, 1 mole should produce 123g.

So, we can use the reactant ratio to compute the product ratio.

123g: 60g

24.6: x g

The actual amount produced of propan-1-ol is 8 g.

In theory, 1 mole should produce 12g.

\frac {Actual Yield}{Theoretical Yield} \times 100%

\frac {8g}{12g} \times 100% = 66.7%

Therefore, the answer is E.