Sodium carbonate weighing 21.2 g reacted with 3.65 g hydrochloric acid. Determine the amount of gas evolved.

univerkov | December 6, 2020 | education

Given:
m (Na2CO3) = 21.2 g
m (HCl) = 3.65 gr
Find: V (CO2) -?
Decision:
1) Na2CO3 + 2HCl = H2O + CO2 + 2NaCl
2) n (Na2CO3) = 21.2gr / 106g / mol = 0.2 mol
Mr (Na2CO3) = 46 + 48 + 12 = 106g / mol
3) n (HCl) = 3.65gr / 36.5g / mol = 0.1 mol
4) n (HCl) is less than n (Na2CO3).
means Na2CO3 in excess
we consider it lacking
5) n (HCl) = 0.5n (CO2) = 0.1 mol
6) from quantity to volume
V (CO2) = 22.4l / mol * 0.1 mol = 2.24l
Answer: V (CO2) = 2.24 liters.

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