Question

A compound M decomposes by two different reactions under different conditions as follows:
$2M\to 3B+C$ $\text{(yield = 80\%) (i)}$
$3M\to D+2C\text{(ii)}$
The total consumed moles of M are 10 and the moles of B and D formed are 6 mol and 1 mol respectively. The $\%$ yield of the (ii) reaction is:

• A. $80\%$
• B. $40\%$
• C. $70\%$
• D. $60\%$

Answer 1

The correct option is D $60\%$

$2M\to 3B+C\text{(yield = 80\%)}$
$3M\to D+2C$
Let us assume the moles of M consumed in reaction (i) to be x mol
2 moles of M give 3 moles of B
x moles of M will give $\frac{3x}{2}$ moles of B
Yield of the reaction is given as 80%
So, moles of B formed $=\frac{80}{100}\times \frac{3x}{2}$
It is given that $=\frac{80}{100}\times \frac{3x}{2}=6$
On solving, x = 5 mol
Moles of M consumed on (ii) reaction $=10–5=5$ mol
For (ii) reaction,
3 moles of M give 1 mol of D
So, 5 mol of M gives $\frac{5}{3}$ moles of D
But moles of M getting formed = 1 mol
So % yield =$\frac{1}{5/3}\times 100$ = 60%