Question

How many grams of $C{l_2}$ gas will be obtained by the complete reaction of $31.5gm$ of potassium permanganate with hydrochloric acid?
[Molar mass of $KMn{O_4}$ = $316gm/mol$]
(A) 71
(B) 17.75
(C) 35.5
(D) 142

Answer 1

Hint: First calculate the number of moles of $KMn{O_4}$ given. Then use the balanced chemical reaction between $KMn{O_4}$ and HCl to check the stoichiometry between $KMn{O_4}$ and $C{l_2}$. Now use the unitary method to calculate the amount of $C{l_2}$ produced from the given amount of $KMn{O_4}$.

Complete step by step solution:
-First of all, we will calculate the number of moles of potassium permanganate involved in this reaction.
The given weight of $KMn{O_4}$ (W) = $31.5gm$
Molar mass of $KMn{O_4}$ given (M) = $316gm/mol$
To calculate the number of moles: $n = \dfrac{W}{M}$
$n = \dfrac{{31.5}}{{316}} = 0.0996$ $mol$
So, the total number of moles of $KMn{O_4}$ involved is $0.0996$ $moles$.
-First of all we will see the reaction involved when potassium permanganate reacts with hydrochloric acid. The reaction is:
$2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2}$
From the reaction we can see that 2 moles of $KMn{O_4}$ will produce 5 moles of $C{l_2}$. We need to calculate the amount of $C{l_2}$ produced when $0.0996$ $moles$ of $KMn{O_4}$ are used. So, for this we will use the unitary method.
2 moles of $KMn{O_4}$ produce → $5$ moles of $C{l_2}$
1 mole of $KMn{O_4}$ produce → $\dfrac{5}{2}$ moles of $C{l_2}$
$0.0996moles$ of $KMn{O_4}$ produce → $\dfrac{5}{2} \times 0.0996$ moles of $C{l_2}$
= $0.249moles$ of $C{l_2}$
So, we can say that $0.249moles$ of $C{l_2}$ are produced.
-We know that the molecular weight of $C{l_2}$ is $ = 2 \times 35.5 = 71gm/mol$
We will now calculate the given weight of $C{l_2}$ using the following equation:
$n = \dfrac{W}{M}$
$0.249 = \dfrac{W}{{71}}$
$W = 0.249 \times 71$
$ = 17.679gm$
So, we now conclude that $17.679gm$ of $C{l_2}$ will be produced.

Hence the correct option will be: (B) 17.75.

Note: In this reaction potassium permanganate ($KMn{O_4}$) being a strong oxidizing agent gets reduced from its +7 state to +2 state and oxidises $HCl$ to $C{l_2}$. This is a redox reaction.