1 mole of mixture of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ requires exactly $28 \mathrm{~g} \mathrm{KOH}$ in…

1 mole of mixture of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ requires exactly $28 \mathrm{~g} \mathrm{KOH}$ in solution for complete conversion of all the $\mathrm{CO}_{2}$ into $\mathrm{K}_{2} \mathrm{CO}_{3}$. How much amount more of $\mathrm{KOH}$ will be required for conversion into $\mathrm{K}_{2} \mathrm{CO}_{3}$ if one mole of mixture is completely oxidized to $\mathrm{CO}_{2}$.
  1. $112 \mathrm{~g}$
  2. $84 \mathrm{~g}$
  3. $56 \mathrm{~g}$
  4. $28 \mathrm{~g}$

Solution

$\mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} ightarrow \mathrm{CO}_{2}$
$\mathrm{CO}_{2}+2 \mathrm{KOH} ightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$
Moles of $\mathrm{KOH}=\frac{28}{56}=0.50$
It corresponds to
$0.25 \mathrm{~mol}$ of $\mathrm{CO}_{2}$
Hence mol of $\mathrm{CO}=1-0.25=0.75 \equiv$ mole of
$\mathrm{CO}_{2}$ formed Mol of KOH requred $=2 \times 0.75=1.5$
$=1.5 \times 56=84 \mathrm{~g}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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