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

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.