Question

1 mole aqueous solution of $S{O}_{2}C{l}_2$ is neutralised by :

• A. $2$ mol of $Ca(OH{)}_2$
• B. $1$ mol of $Ca(OH{)}_2$
• C. $2$ mol of $KOH$
• D. $4$ mol of $KOH$

Answer 1

Answer: (A), (D)

$2$ mol of $Ca(OH{)}_2$
$4$ mol of $KOH$

The reactions are as follows:
$\underset{1mol}{S{O}_{2}C{l}_2}+2H_{2}O⟶\underset{1mol}{H_{2}S{O}_4}+\underset{2mol}{2HCl}$
$1mol{H}_{2}S{O}_4\equiv 1molCa(OH{)}_2\equiv 2molKOH$
$\frac{1}{2}molCa(OH{)}_2\equiv 1molKOH$
$2molHCl\equiv 1molCa(OH{)}_2\equiv 2molKOH$
Thus, mixture of $1$ mol $H_{2}S{O}_4$ and $2$ mol $HCl$ requires $2$ mol $Ca(OH{)}_2$ or $4$ mol of $KOH$.