Question

$100\, mL$ of $N / 5\, HCl$ was added to $1 \,g$ of pure $CaCO _3$ What would remain after the reaction?

• A. 0.5 g of of $CaCO_3$
• B. Neither $CaCO_3$ nor HCI
• C. 50 mL of HCI
• D. 25 mL of HCI

Answer 1

Answer: (B)

No. of moles of $CaCO _3=\frac{w}{M}=\frac{1}{100}$ $=0.01 \,mol$
$\left[\because\right.$ Molar mass of $\left.CaCO _3=100 \,g \,mol ^{-1}\right]$
$100 \, mL$ of $N / 5 HCl =\frac{100}{1000} \times \frac{1}{5}$
$=0.02 \, mol$
The reaction is represented as :
$CaCO _3+2 HCl \longrightarrow CaCl _2+ CO _2+ H _2 O$
$1 $ mole of $CaCO _3$ reacts with 2 moles of $HCl$.
Hence, $0.01$ mole of $CaCO _3$ will react completely with $0.02$ mole of $HCl$.