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Stoichiometric Calculations

Calcium carbo...

Question

Calcium carbonate reacts with aqueous HCl to give $C a C l_{2}$ and $C O_{2}$ according to the reaction given below

$C a C O_{3} (s) + 2 H C l (a q) ⟶ C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)$

What mass of $C a C l_{2}$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of $C a C O_{3}$ ? Name the limiting reagent. Calculate the number of moles of $C a C l_{2}$ formed in the reaction.

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Solution

Molar mass of $C a C O_{3} = 40 + 12 + 3 \times 16 = 100 g m o l^{- 1}$

Moles of $C a C O_{3}$ in 1000 g. $n_{C a C O_{3}} = \frac{Mass (g)}{Molar mass}$

$n_{C a C O_{3}} = \frac{1000 g}{100 g m o l^{- 1}} = 10 m o l M o l a r i t y = \frac{Moles of solute (HCl) \times 1000}{V o l u m e o f s o l u t i o n}$
(It is given that moles of HCl in 250 mL of 0.76 M $H C l = n_{H C l})$
$0.76 = \frac{n_{H C l} \times 1000}{250} n_{H C l} = \frac{0.76 \times 250}{1000} = 0.19 m o l . C a C O_{3} (s) 1 m o l + 2 H C l (a q) 2 m o l ⟶ C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)$

According to the equation,

1 mole of $C a C O_{3}$ reacts with 2 moles HCl

$∴$ 10 moles of $C a C O_{3}$ will react with $\frac{10 \times 2}{1}$ = 20 moles HCl.

But we have only 0.19 moles HCl, so HCl is limiting reagent and it limits the yield of $C a C l_{2}$ .

Since, 2 moles of HCl produces 1 mole of $C a C l_{2}$

0.19 mole of HCl will produce $\frac{1 \times 0.19}{2} = 0.095$ mol $C a C l_{2}$

Molar mass of $C a C l_{2} = 40 + (2 \times 35.5) = 111 g m o l^{- 1}$

$∴$ 0.095 mole of $C a C l_{2} = 0.095 \times 111 = 10.54 g$

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