Question

What mass of $95 \%$ pure $CaCO _{3}$ will be required to neutralise $50 \, mL$ of $0.5\, M HCl$ solution according to the following reaction?
$CaCO _{3( s )}+2 HCl _{( aq )} \rightarrow CaCl _{2( aq )}+ CO _{2( g )}+2 H _{2} O _{( l )}$ [Calculate upto second place of decimal point]

• A. $9.50\, g$
• B. $1.25 \, g$
• C. $1.32\, g$
• D. $3.65 \, g$

Answer 1

Answer: (C)

$CaCO _{3( s )}+2 HCl _{\text {(aq.) }} \rightarrow CaCl _{2 \text { (aq.) }}+ CO _{2( a )}+ H _{2} O _{(0)}$
no. of moles of $CaCO _{3}$ (pure) $=\frac{1}{2} \times$ mole of $HCl$
$ \text { [Mole }=\text { molarity } \times \text { volume(in ltr. })]$
$=\frac{1}{2} \times 0.5 \times \frac{50}{1000}=0.0125$
weight of $CaCO _{3}$ (pure) $=$ mole $\times$ mol. wt
$=0.0125 \times 100=1.25\,g$
$\%$ purity $=\frac{\text { wt. of pure substance }}{\text { wt. of impure sample }} \times 100$
$95=\frac{1.25}{\text { wt. of impure sample }} \times 100$
wt. of impure sample $=\frac{1.25 \times 100}{95}=1.32\, g$