Question

In a series of reactions, 20 mol of L and 10 mol of M were made to react to produce $N$ as a product which was utilized to produce $O$ on reaction with 5 mol of K initially present. Find the moles of $O$ produced.
$4L+3M\to 5N+G$ (Yield = 90 %)
$2N+3K\to 4O+3H$ (Yield = 70 %)

• A. 15.86 mol
• B. 10.52 mol
• C. 6.87 mol
• D. 4.67 mol

Answer 1

The correct option is D 4.67 mol

$4L+3M\to 5N+G$ (Yield = 90 %)
$2N+3K\to 4O+3H$ (Yield = 70 %)
Finding the limiting reagent:
$\text{For L =}\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{20}{4}=5mol\text{for M}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{10}{3}=3.33mol$
Since, the ratio is minimum for M, it will be the limiting reagent.
According to stoichiometry,
3 moles of M produce 5 moles of $N$.
10 moles of M will produce =$\frac{5}{3}\times 10=16.67$ mol
But the yield of the reaction is 90%
Actual amount of $N$ formed = $16.67\times \frac{90}{100}=15molofN$
In reaction (ii),
Finding the limiting reagent:
$\text{For N =}\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{15}{2}=7.5mol\text{For K}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{5}{3}=1.67mol$
So, K is the limiting reagent.
From stoichiometry,
3 moles of K produce 4 moles of $O$.
5 moles of K will produce = $\frac{4}{3}\times 5=6.67molofO$
Actual amount of $O$ formed = $6.67\times \frac{70}{100}=4.67mol$