# Determine the volume of gas formed during the interaction of 0.2 kg and 0.1 l of 5%

July 26, 2021 | education

| **Determine the volume of gas formed during the interaction of 0.2 kg and 0.1 l of 5% hydrochloric acid solution (its density is 0.9 g / cm ^ 3).**

Given:

m (CaCO3) = 0.2 kg = 200 g

V solution (HCl) = 0.1 l = 100 ml

ω (HCl) = 5%

ρ solution (HCl) = 0.9 g / cm3 = 0.9 g / ml

To find:

V (gas) -?

1) CaCO3 + 2HCl => CaCl2 + CO2 ↑ + H2O;

2) n (CaCO3) = m / M = 200/100 = 2 mol;

3) m solution (HCl) = ρ solution * V solution = 0.9 * 100 = 90 g;

4) m (HCl) = ω * m solution / 100% = 5% * 90/100% = 4.5 g;

5) n (HCl) = m / M = 4.5 / 36.5 = 0.12 mol;

6) n (CO2) = n (HCl) / 2 = 0.12 / 2 = 0.06 mol;

7) V (CO2) = n * Vm = 0.06 * 22.4 = 1.3 liters.

Answer: The CO2 volume is 1.3 liters.

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