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Kelvi

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What mass of $95\%$ pure $CaCO_3$ will be required to neutralize $50$ mL of $0.5 M$ HCl solution according to the following reaction? $\\ CaCO_3 (s) + 2HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + 2H_2O (l)$

medium

Equilibrium

2022

chemistry

$1.25$ g

$1.32$ g

$3.65$ g

$9.50$ g

Explanation

To solve this problem, we need to determine the mass of $95\%$ pure $CaCO_3$ required to neutralize $50$ mL of $0.5$ M $HCl. \\$

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The balanced chemical equation is:

\\ CaCO_3 (s) + 2HCl (aq) \rightarrow CaCl_2 (aq) + CO_2 (g) + 2H_2O (l) \\

From the equation, 1 mole of

CaCO_3

reacts with 2 moles of

HCl. \\

Step 1: Calculate the moles of

HCl

present.

\\

Volume of

HCl = 50

mL

= 0.050

L

\\

Molarity of

HCl = 0.5

M

\\

Moles of

HCl =

Molarity

\times

Volume

= 0.5 \times 0.050 = 0.025

moles

\\

Step 2: Calculate the moles of

CaCO_3

required.

\\

From the stoichiometry of the reaction,

\\ 1

mole of

CaCO_3

reacts with 2 moles of

HCl. \\

Moles of

CaCO_3 = \frac{Moles \ of \ HCl}{2} = \frac{0.025}{2} = 0.0125

moles

\\

Step 3: Calculate the mass of pure

CaCO_3

required.

\\

Molar mass of

CaCO_3 = 40.08 \, (

Ca

) + 12.01 \, (

C

) + 3 \times 16.00 \, (

O

) = 100.09 \,

g/mol

\\

Mass of pure

CaCO_3 =

Moles

\times

Molar mass

= 0.0125 \times 100.09 = 1.251125 \,

g

\\

Step 4: Calculate the mass of

95\%

pure

CaCO_3

required.

\\

Let

x

be the mass of

95\%

pure

CaCO_3. \\ 0.95x = 1.251125 \\ x = \frac{1.251125}{0.95} = 1.3175 \,

g

\\

Therefore, the mass of

95\%

pure

CaCO_3

required is approximately

1.32 \,

g.

\\

This corresponds to Option 2.