Question

A compound M decomposes in two different ways under different conditions as follows:
$3M\to 2B+C$ $\text{(yield = 70 \%) (i)}$
$5M\to 5D+2C\text{(ii)}$
The total consumed moles of M are 9 and the moles of B and D formed are 2 mol and 3 mol respectively. The % yield of the (ii) reaction is:

• A. 80%
• B. 75%
• C. 70%
• D. 65%

Answer 1

The correct option is B 75%

$3M\to 2B+C\text{(yield = 60\%)}$
$5M\to 5D+2C$

Let us assume the moles of M consumed in the reaction (i) to be x mol
3 moles of M give 2 moles of B.
x moles of M will give $\frac{2x}{3}$ moles of B.
Yield of the reaction is given as 60%
So, the moles of B formed $=\frac{60}{100}\times \frac{2x}{3}$
It is given that, $\frac{60}{100}\times \frac{2x}{3}=2$
On solving, x = 5 mol
Moles of M consumed in (ii) reaction $=9–5=4$ mol
For (ii) reaction,
5 moles of M give 5 mol of D.
So, 4 mol of M gives $4$ moles of D.
But, moles of D getting formed = 3 mol
So % yield =$\frac{3}{4}\times 100$ = 75%