Na₂CO₃ + 2HCI → 2NaCI + H₂O + CO₂. Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid?
(1 mole of gas occupies 22.4 dm³ at s.t.p)
(Na = 23, C = 12, O = 16)

44.8 dm³

11.2 dm³

100.1 dm³

3.0 dm³

22.4 dm³

Correct answer is B

Na₂CO₃ + 2HCI → 2NaCI +H₂O + CO₂
53 gm Na₂CO₃ will therefore liberate (22.4)/(106) x (53)/(1) = 11.2 dm³

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Na2CO3 + 2HCI → 2NaCI + H2O + CO2. Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid? (1 mole of gas occupies 22.4 dm3 at s.t.p) (Na = 23, C = 12, O = 16)