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$\mathrm{LiOH}$ reacts with $\mathrm{CO}_{2}$ to form $\mathrm{Li}_{2} \mathrm{CO}_{3}$ (atomic mass of $\mathrm{Li}=7$ ). The amount of $\mathrm{CO}_{2}$ (in g) consumed by $1 \mathrm{~g}$ of LiOH is closest to
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The correct answer is:
$0.916$
$\begin{array}{l}
2 \mathrm{LiOH}+\mathrm{CO}_{2} \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \\
\frac{1}{24} \frac{1}{24 \times 2}
\end{array}$
No. of moles of $\mathrm{CO}_{2}=\frac{1}{48}$
mass of $\mathrm{CO}_{2}=\frac{1}{48} \times 44=0.916 \mathrm{~g}$
2 \mathrm{LiOH}+\mathrm{CO}_{2} \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \\
\frac{1}{24} \frac{1}{24 \times 2}
\end{array}$
No. of moles of $\mathrm{CO}_{2}=\frac{1}{48}$
mass of $\mathrm{CO}_{2}=\frac{1}{48} \times 44=0.916 \mathrm{~g}$
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