JAMB Chemistry 2019 Objective — Question 11

Question 11 of 40 from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2019 Objective paper, with the correct answer and a full explanation.

If 15g of impure Na₂CO₃ reacted with excess HNO₃ and 4.4g of CO₂ was produced, what is the percentage purity of the Na₂CO₃? (Na=23, C=12, O=16)

A. 35.3%
B. 10.0%
C. 70.667%Correct
D. 90.0%

Explanation

Na₂CO₃ + 2HNO₃ → 2NaNO₃ + H₂O + CO₂. n(CO₂) = 4.4/44 = 0.1mol. Mole ratio Na₂CO₃:CO₂ = 1:1, so n(Na₂CO₃) = 0.1mol. Molar mass Na₂CO₃ = 106g/mol, so mass of pure Na₂CO₃ = 0.1×106 = 10.6g. % purity = 10.6/15 × 100 = 70.667%.

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