The volume of carbon dioxide generated by the interaction of 10 g of calcium carbonate with
The volume of carbon dioxide generated by the interaction of 10 g of calcium carbonate with a solution containing 36.5 g of hydrochloric acid is equal to.
Let’s find the amount of acid substance by the formula:
n = m: M.
M (HCl) = 35 + 1 = 36 g / mol.
n = 36.5 g: 36 g / mol = 1.01 mol.
Find the amount of calcium carbonate substance:
M (CaCO3) = 100 g / mol.
n = 10 g: 100 g / mol = 0.1 mol.
Hydrochloric acid is given in excess.
The amount of carbon dioxide is calculated by the deficiency (calcium carbonate).
Let’s compose the reaction equation, find the quantitative relationships of substances
CaCO3 + 2HCl = CaCl2 + CO2 ↑ + H2O.
According to the reaction equation, there is 1 mole of gas per mole of calcium carbonate. Substances are in quantitative ratios 1: 1.
The amount of substance will be the same.
n (CO2) = n (CaCO3) = 0.1 mol.
Let’s find the volume of carbon dioxide.
V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.
V = 0.1 mol × 22.4 L / mol = 2.24 L.
Answer: 2.24 liters.