Some CO_2 gas was kept in a sealed container | vedclass

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(A) The reaction between $CO_2$ and $Ca(OH)_2$ is: $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$. Let the moles of $CO_2$ be $n$. According to the stoic

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) The reaction between $CO_2$ and $Ca(OH)_2$ is: $Ca(OH)_2 + CO_2 ightarrow CaCO_3 + H_2O$. Let the moles of $CO_2$ be $n$. According to the stoichiometry,$n$ moles of $CO_2$ react with $n$ moles of $Ca(OH)_2$. Given that this amount of $CO_2$ reacted with exactly half the initial amount of $Ca(OH)_2$,the total initial moles of $Ca(OH)_2 = 2n$. Excess $Ca(OH)_2 = 2n - n = n$. This excess $Ca(OH)_2$ is neutralized by $HCl$: $Ca(OH)_2 + 2HCl ightarrow CaCl_2 + 2H_2O$. Equivalents of $Ca(OH)_2 = $ Equivalents of $HCl$. $n 	imes 2 = 0.1 	imes 0.040 	imes 1$. $2n = 0.004 \implies n = 0.002 \ mol$. At $STP$ $(1 \ atm, 273 \ K)$,$1 \ mol$ of gas occupies $22400 \ cm^3$. Volume $x = 0.002 	imes 22400 = 44.8 \ cm^3$. Rounding to the nearest integer,$x = 45 \ cm^3$.

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