What mass of 95% pure will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? [Calculate upto second place of decimal point.] [2022]

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Solution

Q.
What mass of 95% pure ${CaCO}_{3}$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction

${CaCO}_{3 (s)} + 2 {HCl}_{(a q)} \to {CaCl}_{2 (a q)} + {CO}_{2 (g)} + 2 H_{2} O_{(l)}$

[Calculate upto second place of decimal point.] [2022]

1 1.25 g

2 1.32 g

3 3.65 g

4 9.50 g

Ans.
(2)

Volume of HCl = 50 mL = 0.05 L

Molarity of HCl = 0.5 M

$∴$ Moles of HCl = $0.05 \times 0.5 = 0.025$ moles

$C a C O_{3} + 2 H C l ⟶ C a C l_{2} + C O_{2} + H_{2} O$

For 2 moles of HCl, $C a C O_{3}$ required = 1 mole

$∴$ For 0.025 moles of HCl, $C a C O_{3}$ required = $\frac{0.025}{2}$ moles

Mass of $C a C O_{3}$ required = $100 \times \frac{0.025}{2} = 1.25$ g

For 95% pure $C a C O_{3}$ , mass of $C a C O_{3}$ required = $\frac{1.25}{95} \times 100 g = 1.315 g \approx 1.32$ $g$

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