If 40 g of CaCO₃ is treated with 40 g of HCl, which of the reactants will act as limiting reagent?

$\underset{(100 \mathrm{g})}{\underset{1 \mathrm{mole}}{{\mathrm{CaCO}}_3}}+\underset{(73 \mathrm{g})}{\underset{2 \mathrm{moles}}{2\mathrm{HCl}}}\to {\mathrm{CaCl}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$
100 g of CaCO₃ reacts with 73 g of HCI
40 g of CaCO₃ will react with $\frac{73}{100}\times 40=29.2 \mathrm{g} \mathrm{of} \mathrm{HCl}$
Since CaCO₃ is completely consumed and some amount (40 - 29.2 = 10.8 g) of HCI remains unreacted and hence, CaCO₃ is the limiting reagent.