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%
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.