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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