When sodium carbonate with a mass of 10 g interacts with an excess of nitric acid, a gas with a volume of?

univerkov | January 17, 2021 | education

Given:
m (Na2CO3) = 10 g
To find:
V (CO2)
Decision:
Na2CO3 + 2HNO3 = 2NaNO3 + H2O + CO2
n (Na2CO3) = m / M = 10 g / 106 g / mol = 0.09 mol
n (Na2CO3): n (CO2) = 1: 1
n (CO2) = 0.09 mol
V (CO2) = n * Vm = 0.09 mol * 22.4 L / mol = 2.016 L
Answer: 2,016 l

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