Answer to Question #255941 in Chemistry for Folami

Question

A is a solution of hydrochloric acid containing 3.60g of Hcl in 1dm³ of solution. B is a solution of impure of sodium carbonate.25cm³ of B require 22.7cm³ of A for complete reaction.Given that 5.00g of impure B was dissolved in 1dm³ solution,what is the percentage by mass of pure sodium carbonate in the impure sample​

Answer 1

Solution:

The molar mass of HCl is 36.46 g/mol

The molar mass of Na₂CO₃ is 105.99 g/mol

Calculate the molarity of solution A:

(3.60 g HCl) × (1 mol HCl / 36.46 g HCl) × (1 / 1dm³) × (1 dm³ / 1 L) = 0.09874 mol HCl dm^-3

C_M(HCl) = 0.09874 mol/dm³

Balanced chemical equation:

Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)

According to stoichiometry:

n(Na₂CO₃) = n(HCl) / 2

2 × C_M(Na₂CO₃) × V(Na₂CO₃) = C_M(HCl) × V(HCl)

2 × C_M(Na₂CO₃) × (25 cm³) = (0.09874 mol/dm³) × (2 × 22.7 cm³)

C_M(Na₂CO₃) = (0.09874 mol/dm³ × 22.7 cm³) / (2 × 25 cm³) = 0.04483 mol/dm³

C_M(Na₂CO₃) = 0.04483 mol/dm³

Calculate the moles of Na₂CO₃:

(0.04483 mol Na₂CO₃ / 1 dm³) × (1 dm³) = 0.04483 mol Na₂CO₃

Calculate the mass of Na₂CO₃:

(0.04483 mol Na₂CO₃) × (105.99 g Na₂CO₃ / 1 mol Na₂CO₃) = 4.75 g Na₂CO₃

Mass of pure Na₂CO₃ is 4.75 grams

Calculate the percentage by mass of pure Na₂CO₃ in the impure sample:

%Na₂CO₃ = (Mass of pure Na₂CO₃ / Mass of impure sample) × 100%

Mass of impure sample is 5.00 grams

Therefore,

%Na₂CO₃ = (4.75 g / 5.00 g) × 100% = 95%

%Na₂CO₃ = 95%

Answer: The percentage by mass of pure Na₂CO₃ in the impure sample​ is 95%