Question

How many litres of $C{l}_2$ at STP will be liberated by the oxidation of $NaCl$ with $10$ g $KMn{O}_4$?

• A. $3.54$ L
• B. $7.08$ L
• C. $1.77$ L
• D. None of these

Answer 1

The correct option is A $3.54$ L

The reaction is as follows:
$NaCl+KMn{O}_4\to C{l}_2+M{n}^{+2}$
At equivalent point,
$N_1{V}_1=N_2{V}_2$
Number of equivalents of $KMn{O}_4$ used is $\frac{10\times 5}{158}=0.316$.
So, volume of $C{l}_2$ produces is $0.361\times 11.2=3.54$ L.