What mass of 80% pure $CaC{O}_3$ will be required to neutralize 50 mL of 0.5M HCl solution.

$CaC{O}_3+2HCl\to CaC{l}_2+C{O}_2+2H_{2}O$

No.of moles of HCl = M.V (L)

= 0.5 x $\frac{50}{1000}$

= 0.025 moles.

1 mole of $CaC{O}_3$ requires  2 moles of HCl

100g of $CaC{O}_3\to$ 2 moles of HCl

              ?              $\to$by 0.025g of HCl

$=\frac{100\times 0.025}{2}=1.25g$

Mass of 80% pure CaCO₃

=$\frac{1.25\times 100}{80}=1.56g$