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
The correct answer is: B
Explanation
Na2CO3 + 2HCI → 2NaCI +H2O + CO253 gm Na2CO3 will therefore liberate (22.4)/(106) x (53)/(1) = 11.2 dm3