Question

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

Answer 1

Reaction is as follows:
$NaCl+KMn{O}_4\to C{l}_2+M{n}^{+2}$
At equivalent point,
$N_1{V}_1=N_2{V}_2$
No. of equivalents of $KMn{O}_4$ used is $10\times \left(\frac{5}{158}\right)=0.316$.
So, volume of $C{l}_2$ produces is $0.361\times 11.2=3.54$ L.
So, integer value is $3$.