Question

$1$ mole of mixture of $CO$ and $C{O}_2$ requires exactly $28$g $KOH$ in solution for complete conversion of all the $C{O}_2$ into $K_{2}C{O}_3$ . How much amount more of $KOH$ will be required for conversion into $K_{2}C{O}_3$ if one mole of mixture is completely oxidized to $C{O}_2$.

• A. 112 g
• B. 84 g
• C. 56 g
• D. 28 g

Answer 1

The correct option is B 84 g

$CO+\frac{1}{2}{O}_2\to C{O}_2$

$C{O}_2+2KOH\to K_{2}C{O}_3+H_{2}O$

Moles of KOH $=\frac{28}{56}=0.50$.

lt corresponds to 0.25 moles of $C{O}_2$.

Hence, moles of $CO=1-0.25=0.75moles$ of $C{O}_2$ formed.

Moles of KOH required $=$ $2\times 0.75=1.5$

mass of KOH required $=1.5\times 56=84g$.
84 gm more KOH required.

Therefore, option B is correct.