How many liters of CO2 are released when 10 grams of Ca CO3 interacts with HCE chloride acid

univerkov | September 30, 2021 | education

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

According to the equation: 1 mol of CaCO3 enters into the reaction. Let’s find its molar mass:

M (CaCO3) = 40 + 12 + 16 * 3 = 100 g / mol

According to the equation: 1 mol of CO2 is formed, i.e. 22.4 l

We make the proportion:

10 g CaCO3 – xl CO2

100 g CaCO3 – 22.4 l CO2

Hence, x = 10 * 22.4 / 100 = 2.24 liters of CO2.

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