Determine the volume of gas that will be released during the interaction of 200 g of 10% hydrochloric

univerkov | November 25, 2020 | education

Determine the volume of gas that will be released during the interaction of 200 g of 10% hydrochloric acid HCl with potassium carbonate K2Co3.

2HCl + K2CO3 = 2KCl + CO2 + H2O
m (HCl) pure = 200g * 0.1 = 20g
n (HCl) = m (HCl) / M (HCl) = 20g / 36.5 g / mol = 0.548 mol approx
n (CO2) = n (hcl) / 2 = 0.548 / 2 = 0.274 mol
V (CO2) = n (CO2) * Vm = 0.274 mol × 22.4 L / mol = 6.14 L approximately
Answer: 6.14 L

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