# What mass of CaCo3 is required to obtain Co2 volume = 11.2 liters?

August 30, 2021 | education

| Let’s write down the solution:

1. According to the condition of the problem, we compose the equation:

X g -? V = 11.2 liters.

CaCO3 = CaO + CO2 – decomposition, carbon monoxide obtained (4);

2. Let’s make the calculations:

M (CaCO3) = 70 g / mol.

M (CO2) = 44 g / mol.

3. Proportion:

1 mol of gas at normal level – 22.4 liters.

X mol (CO2) -11.2 l. hence, X mol (CO2) = 1 * 11.2 / 22.4 = 0.5 mol.

Y (CaCO3) = 0.5 mol since the amount of substances is 1 mol.

4. Find the mass of the original substance:

m (CaCO3) = Y * M = 0.5 * 70 = 35 g.

Answer: The mass of calcium carbonate is 35 g.

One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.