$\mathrm{x} \mathrm{g}$ of methane was burnt completely in the presence of oxygen. The liberated gases were…
$\mathrm{x} \mathrm{g}$ of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing $370 \mathrm{~g}$ of $\mathrm{Ca}(\mathrm{OH})_2$. The weight of white precipitate obtained was 500 . What is the value of $x$ (in g)?
(Given : $\mathrm{C}=12 ; \mathrm{H}=1 ; \mathrm{Ca}=40 ; \mathrm{O}=16 \mathrm{u}$ )
16
80
160
120
Solution
$\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
Thus, 1 mole of $\mathrm{CH}_4$ gives 1 mole of $\mathrm{CO}_2$ gas. $\mathrm{CO}_2$ gas combines with $\mathrm{Ca}(\mathrm{OH})_2$ as :-
$
\mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}
$
Thus, 1 mole of $\mathrm{Ca}(\mathrm{OH})_2$ gives 1 mole of $\mathrm{CaCO}_3$ precipitate or $74 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_2$ gives $100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ precipitate.
Therefore, $500 \mathrm{~g}$ of precipitate would be obtained by : $74 \times 5=370 \mathrm{~g}$ or 5 moles of $\mathrm{Ca}(\mathrm{OH})_2$.
Since products are formed as per the stoichiometric coefficients, there would be 5 moles of $\mathrm{CO}_2$ released in the first reaction. 5 moles of $\mathrm{CO}_2$ would be released by 5 moles or $5 \times 16=80 \mathrm{~g}_{\text {of }} \mathrm{CH}_4$.
Thus, $x=80$