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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3containing equimolar amounts of both - Chemistry

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3containing equimolar amounts of both?

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Solution

Let the amount of Na2CO3 in the mixture be x g.

Then, the amount of NaHCO3 in the mixture is (1 − x) g.

Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16

= 106 g mol−1

∴ Number of moles Na2CO3 = `x/106` mol

Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16

= 84 g mol−1

∴ Number of moles of NaHCO3 = `(1-x)/84` mol

According to the question,

`x/106 = (1-x)/84`

⇒ 84x = 106 − 106x

⇒ 190x = 106

⇒ x = 0.5579

Therefore, number of moles of Na2CO3 = `0.5579/106` mol

= 0.0053 mol

And, number of moles of NaHCO= `(1-0.5579)/84`

 = 0.0053 mol

HCl reacts with Na2CO3 and NaHCO3 according to the following equation.

1 mol of Na2CO3 reacts with 2 mol of HCl.

Therefore, 0.0053 mol of Na2CO3 reacts with 2 × 0.0053 mol = 0.0106 mol.

Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl.

Therefore, 0.0053 mol of NaHCO3 reacts with 0.0053 mol of HCl.

Total moles of HCl required = (0.0106 + 0.0053) mol

= 0.0159 mol

In 0.1 M of HCl,

0.1 mol of HCl is preset in 1000 mL of the solution.

Therefore, 0.0159 mol of HCl is present in = `(1000xx0.0159)/0.1` mol

= 159 mL of the solution

Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na2CO3 and NaHCO3,containing equimolar amounts of both.

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 12 Chemistry Textbook
Chapter 2 Solutions
Q 6 | Page 60
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