At STP ' $x$ ' $g$ of a metal hydrogen carbonate $\left(\mathrm{MHCO}_3\right)$ (molar mass $84 \mathrm{~g}…
At STP ' $x$ ' $g$ of a metal hydrogen carbonate $\left(\mathrm{MHCO}_3\right)$ (molar mass $84 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on heating gives $\mathrm{CO}_2$, which can completely react with 0.2 moles of MOH (molar mass 40 $\mathrm{g} \mathrm{mol}^{-1}$ ) to give $\mathrm{MHCO}_3$. The value of ' $x$ ' is
$67.2$
$33.6$
$11.2$
$22.4$
Solution
Given, 0.2 moles of MOH reacts completely with $\mathrm{CO}_2$
$\mathrm{CO}_2+\mathrm{MOH} \rightarrow \mathrm{M} \mathrm{HCO}_3$
$0.2 \quad 0.2$ moles moles
Now, $2 \mathrm{MHCO}_3 \xrightarrow{\Delta} \mathrm{M}_2 \mathrm{CO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$
1 mole of $\mathrm{CO}_2$ is obtained from 2 moles of $\mathrm{MHCO}_3$
0.2 mole of $\mathrm{CO}_2$ will be obtain from $=2 \times 0.2$ $=0.4$ mole of $\mathrm{MHCO}_3$
Hence, amount ' x ' $=0.4 \times 84=33.6 \mathrm{~g}$