Question

When $1g$ of $Ca{CO}_3$ reacts with $50ml$ of $0.1M$ $HCl$, the volume of ${CO}_2$ produced is:

• A. $11.2mL$
• B. $22.4mL$
• C. $112mL$
• D. $56mL$

Answer 1

The correct option is D $56mL$

${CaCO}_3+2HCl\to {CaCl}_2+{CO}_2+H_{2}O$

$1$ gm of ${CaCO}_3=\frac{1}{100}=0.01$ moles

$50$ml of $01$ M HCl\equiv \dfrac { 0.1 }{ 20 } $moles$=0.005$ moles

$\therefore$ no. of moles of ${CO}_2$ produced $=\frac{0.005}{2}$ moles

$=0.0025$ moles

Now, $1$ mole ${CO}_2\equiv 22.4$ litre (at STP)

$\Rightarrow 0.0025moles\equiv \frac{22.4\times 0.0025}{1}=0.056litre=56ml.$