Excess of carbon dioxide is passed through 50 | vedclass

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(C) The reaction of $CO_{2}$ with $Ca(OH)_{2}$ in excess leads to the formation of $CaCO_{3}$ and water: $Ca(OH)_{2} + CO_{2} \rightarrow CaCO_{3} + H

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(C) The reaction of $CO_{2}$ with $Ca(OH)_{2}$ in excess leads to the formation of $CaCO_{3}$ and water: $Ca(OH)_{2} + CO_{2} ightarrow CaCO_{3} + H_{2}O$. Millimoles of $Ca(OH)_{2} = 50 \ mL 	imes 0.5 \ M = 25 \ mmol$. Since the stoichiometry is $1:1$,millimoles of $CaCO_{3}$ formed $= 25 \ mmol$. The reaction of $CaCO_{3}$ with $HCl$ is: $CaCO_{3} + 2HCl ightarrow CaCl_{2} + H_{2}O + CO_{2}$. The milliequivalents of $CaCO_{3}$ $= 	ext{millimoles} 	imes n	ext{-factor} = 25 	imes 2 = 50 \ meq$. At neutralization,milliequivalents of $HCl = 	ext{milliequivalents of } CaCO_{3} = 50 \ meq$. Volume of $0.1 \ N \ HCl = \frac{50 \ meq}{0.1 \ N} = 500 \ cm^{3}$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

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