Question

A decomposes under different environment separately to form B and C according to reactions:
$A\to 3B\dots \text{(i)}$
$A\to 2C\dots \text{(ii)}$
The yield of reaction (i) and (ii) is found to be 60% and 40% respectively and amount of B and C is found to be $20\text{g}$ and $15\text{g}$ respectively. Calculate the total amount of A which were taken initially.
(Molar mass of $A=30\text{g/mol}$)

• A. $18.5\text{g}$
• B. $25.5\text{g}$
• C. $70.8\text{g}$
• D. $60.0\text{g}$

Answer 1

The correct option is C $70.8\text{g}$

$A\to 3B\dots \text{(i)}$
$A\to 2C\dots \text{(ii)}$
Since, the molar mass of A is given to be 30 g/mol and 1 mol of A produces 3 mol of B and 2 mol of C.
From the law of conservation of mass we know,
Molar mass of B $=\frac{30}{3}=10\text{g/mol}$
Molar mass of C $=\frac{30}{2}=15\text{g/mol}$
Number of moles of B produced $=\frac{\text{Weight produced}}{\text{Molar mass}}=\frac{20}{10}=2\text{mol}$
Since the yield of the reaction (i) is 60 %
Moles of A taken in the reaction (i) $=\frac{2}{0.6}\times \frac{1}{3}=1.11\text{mol}$
Number of moles of C produced $=\frac{\text{Weight produced}}{\text{Molar mass}}=\frac{15}{15}=1\text{mol}$
Moles of A taken in the reaction (ii) $=\frac{1}{0.4}\times \frac{1}{2}=1.25\text{mol}$
Total amount of A taken = Total number of moles of A $\times$ Molar mass of A
$=(1.11+1.25)\times 30=70.8\text{g}$
Therefore, the total amount of A initially taken is $70.8\text{g}$