Question

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

(Atomic weight: $Mn=55$ and $K=39$)

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

Answer 1

Answer: (A)

$3.54$

$10NaCl+8H_2{SO}_4+2{KMnO}_4⟶{5Cl}_2+2{MnSO}_4+5{Na}_2{SO}_4+K_2{SO}_4+8H_{2}O$

$2\times 158=316g=5\times 22.4L$ at $STP$

$\therefore$ $10g$ of ${KMnO}_4=\frac{5\times 22.4\times 10}{316}=3.5L$