# What volume of carbon dioxide will be released when 50 g of marble interacts with an excess of hydrochloric

May 29, 2021 | education

| **What volume of carbon dioxide will be released when 50 g of marble interacts with an excess of hydrochloric acid, if the practical yield of carbon dioxide is 95%?**

Let’s write the reaction equation:

CaCO3 + 2HCl = CACl + H2O + CO2

m (CaCO3) / M (CaCO3) = V (CO2) / Vm (CO2)

Vm (CO2) = 22.4 l / mol.

Let’s calculate the molar mass of marble:

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

Let’s determine the volume of carbon dioxide:

V (CO2) = m (CaCO3) * Vm (CO2) / M (CaCO3)

Substitute the numerical values:

V (CO2) = 50 * 22.4 / 100 = 11.2 liters.

Let’s make the proportion:

11.2 l – 100%

x – 95%

Then, expressing from the proportion x, we get:

x = 11.2 * 95/100 = 10.64 liters.

Answer: 10.64 liters of carbon dioxide will be released.

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