# Find the volume of carbon dioxide and the amount of calcium chloride that is formed

Find the volume of carbon dioxide and the amount of calcium chloride that is formed by the interaction of 15 g of calcium carbonate with hydrochloric acid.

1. Let’s write down the reaction equation:

CaCO3 + 2HCl = CaCl2 + CO2 + H2O.

2. Find the amount of reacted calcium carbonate:

n (CaCO3) = m (CaCO3) / M (CaCO3) = 15 g / 100 g / mol = 0.15 mol.

3. According to the reaction equation, we find the amount of calcium chloride, as well as the volume of carbon dioxide (Vm – molar volume, constant equal to 22.4 l / mol):

n (CaCl2) = n (CaCO3) = 0.15 mol.

n (CO2) = n (CaCO3) = 0.15 mol.

V (CO2) = n (CO2) * Vm = 0.15 mol * 22.4 l / mol = 3.36 l.

Answer: n (CaCl2) = 0.15 mol; V (CO2) = 3.36 l.

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