Question

Calcium carbonate reacts with aqueous $\mathrm{HCl}$to give ${\mathrm{CaCl}}_2$ and ${\mathrm{CO}}_2$according to the reaction given below

$\underset{\mathrm{Calcium}\mathrm{Carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{aq}\right)}⟶\underset{\mathrm{Calcium}\mathrm{Chloride}}{{\mathrm{CaCl}}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

What mass of ${\mathrm{CaCO}}_3$ is required to react completely with 25 mL of 0.75 M HCl?

1) 0.6844

2) 0.9375

3) 0.4265

4) 0.2785

Answer 1

The given equation is
$\underset{\mathrm{Calcium}\mathrm{Carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{aq}\right)}⟶\underset{\mathrm{Calcium}\mathrm{Chloride}}{{\mathrm{CaCl}}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Number of moles of HCl = Molarity x Volume
$=0.75\mathrm{M}\times 0.025\mathrm{L}=0.01875\mathrm{mol}$
As, the mole ratio between ${\mathrm{CaCO}}_3$ and HCl is 1:2. So, the number of moles of ${\mathrm{CaCO}}_3$ should be half of the number of moles of HCl
Number of moles of ${\mathrm{CaCO}}_3$$=\frac{1}{2}\times 0.01875\mathrm{mol}=0.009375\mathrm{mol}$
Molar mass of ${\mathrm{CaCO}}_3$$=100\mathrm{g}/\mathrm{mol}$
Mass of Calcium Carbonate = Moles$\times$Molar mass $=0.009375\mathrm{mol}\times 100\mathrm{g}/\mathrm{mol}=0.9375\mathrm{g}$
So option (2) is correct