Ncert Exercises & Solutions

43. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction given below

${CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + {CO}_{2} (g) + H_{2} O (l)$

What mass of CaCl₂ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃? Name the limiting reagent. Calculate the number of moles of CaCl₂ formed in the reaction.

Molar mass of

{CaCO}_{3} = 40 + 12 + 3 \times 16 = 100 g {mol}^{- 1}

Moles of

{CaCO}_{3} in 1000 g, n_{{CaCO}_{3}} = \frac{Mass (g)}{Molar mass}

n_{{CaCO}_{3}} = \frac{1000 g}{100 g {mol}^{- 1}} = 10 mol

Molarity = \frac{Moles of solute (HCl) \times 1000}{Volume of solution}

(It is given that moles of HCl in 250 mL of 0.76 M HCl = n_{HCl})

0.76 = \frac{n_{HCl} \times 1000}{250}

n_{HCl} = \frac{0.76 \times 250}{1000} = 0.19 mol .

\underset{1 mol}{{CaCO}_{3} (s)} + \underset{2 mol}{2 HCl (aq)} \to {CaCl}_{2} (aq) + {CO}_{2} (g) + H_{2} O (l)

According to the equation,

1 mole of

{CaCO}_{3}

reacts with 2 moles of HCl

∴

10 moles of

{CaCO}_{3}

will react with

\frac{10 \times 2}{1} = 20

moles HCl.

But we have only 0.19 moles HCl, so HCl is limiting reagnt and it limits the yield of CaCl₂.

Since, 2 moles of HCl produce 1 mole of CaCl₂

0.19 mole of HCl will produce

\frac{1 \times 0.19}{2} = 0.095 mol {CaCl}_{2}

Molar mass of

{CaCl}_{2} = 40 + (2 \times 35.5) = 111 g {mol}^{- 1}

∴

0.095 mole of

{CaCl}_{2} = 0.095 \times 111 = 10.54 g

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