Question

If $40 \,g$ of $CaCO_{3}$ is treated with $40\, g$ of $HCl$, which of the reactants will act as limiting reagent?

• A. $CaCO_3$
• B. $HCl$
• C. Both (a) and (b)
• D. None of these

Answer 1

Answer: (A)

$CaCO_3$

Answer 2

$\underset{\text{1mole(100g)}}{ {CaCO3}}$+$\underset{\text{2moles(73g)}}{ {HCl}}$ ${->}$ $CaCl_{2}+H_{2}O+CO_{2}$ $100\, g$ of $CaCO_3$ reacts with $73\, g$ of $HCl$ $40\, g$ of $CaCO_3$ will react with $\frac{73}{100} \times 40 = 29.2\,g$ of $HCl$ Since $CaCO_3$ is completely consumed and some amount $(40 - 29.2 = 10.8\, g)$ of $HCl$ remains unreacted and hence, $CaCO_{3}$ is limiting reagent.