What volume of carbon dioxide will be released when 0.5 mol HCL interacts with Na2CO3?

univerkov | January 17, 2021 | education

Given:
n (HCl) = 0.5 mol
To find:
V (CO2)
Decision:
Na2CO3 + 2HCl = 2NaCl + H2O + CO2
n (HCl): n (CO2) = 2: 1
n (CO2) = 0.5 mol / 2 = 0.25 mol
V (CO2) = n * Vm = 0.25 mol * 22.4 l / mol = 5.6 l
Answer: 5.6 L

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