JAMB - Chemistry (1981 - No. 23)
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)
(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
Selitys
Na2CO3 + 2HCI → 2NaCI +H2O + CO2
53 gm Na2CO3 will therefore liberate (22.4)/(106) x (53)/(1) = 11.2 dm3
53 gm Na2CO3 will therefore liberate (22.4)/(106) x (53)/(1) = 11.2 dm3