What volume of carbon dioxide will be released during the decomposition of 300 g of MgCO3 with 15% impurities?

1. Let’s write the reaction equation:
MgCO3 = MgO + CO2.
2. Find the amount of magnesium carbonate:
ω (MgCO3) = 100% – ω (impurities) = 100% – 15% = 85%.
m (MgCO3) = (m (mixture) * ω (MgCO3)) / 100% = 300 g * 85% / 100% = 255 g.
n (MgCO3) = m (MgCO3) / M (MgCO3) = 255 g / 84 g / mol = 3.04 mol.
3. According to the reaction equation, we find the amount and then the volume of carbon dioxide (Vm – molar volume, constant equal to 22.4 l / mol):
n (CO2) = n (MgCO3) = 3.04 mol.
V (CO2) = n (CO2) * Vm = 3.04 mol * 22.4 l / mol = 68 l.
Answer: V (CO2) = 68 l.