How many mL of 0.1 M HCl are required to react completely with 1 g mixture of

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

$\therefore$ Number of moles $\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{x}{106} \mathrm{~mol}$

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

= 84 g mol−1

$\therefore$ Number of moles of $\mathrm{NaHCO}_{3}=\frac{1-x}{84} \mathrm{~mol}$

According to the question,

$\frac{x}{106}=\frac{1-x}{84}$

⇒ 84x = 106 − 106x

⇒ 190x = 106

⇒ x = 0.5579

Therefore, number of moles of $\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{0.5579}{106} \mathrm{~mol}$

= 0.0053 mol

And, number of moles of $\mathrm{NaHCO}_{3}=\frac{1-0.5579}{84}$

= 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, $0.0159 \mathrm{~mol}$ of $\mathrm{HCl}$ is present in $\frac{1000 \times 0.0159}{0.1} \mathrm{~mol}$

= 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.