28 g KOH is required to completely neutraliz | vedclass

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Question

(A) The reaction for heating $CaCO_3$ is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$.Neutralization of $CO_2$ by $KOH$ is: $2KOH + CO_2 \rightarrow K_2C

Vedclass · Accepted Answer

$41.6$

Vedclass · Answer

(A) The reaction for heating $CaCO_3$ is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$.Neutralization of $CO_2$ by $KOH$ is: $2KOH + CO_2 ightarrow K_2CO_3 + H_2O$.From the stoichiometry, $2 \ mol$ of $KOH$ reacts with $1 \ mol$ of $CO_2$, which is produced from $1 \ mol$ of $CaCO_3$.Thus, $2 \ mol$ of $KOH$ reacts with $1 \ mol$ of $CaCO_3$.$2 	imes 56 \ g$ of $KOH$ reacts with $100 \ g$ of $CaCO_3$.$112 \ g$ of $KOH$ reacts with $100 \ g$ of $CaCO_3$.Therefore, $28 \ g$ of $KOH$ reacts with $\frac{100 	imes 28}{112} = 25 \ g$ of pure $CaCO_3$.Percentage purity $= \frac{	ext{mass of pure } CaCO_3}{	ext{mass of impure } CaCO_3} 	imes 100 = \frac{25}{60} 	imes 100 = 41.66\% \approx 41.6\%$.

$28 \ g$ $KOH$ is required to completely neutralize $CO_2$ produced on heating $60 \ g$ of impure $CaCO_3$. The percentage purity of $CaCO_3$ is approximately (molar masses of $KOH$ and $CaCO_3$ are $56$ and $100 \ g \ mol^{-1}$, respectively).

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