How many liters of CO2 will be released when 294 g of magnesium carbonate is ignited?

Let’s write the reaction equation for the solution:
MgCO3 = MgO + CO2 – the decomposition reaction of magnesium carbonate, carbon dioxide is released;
Calculate M (MgCO3) = 24.3 + 12 + 48 = 84.3 g / mol
Determine the number of mol of MgCO3:
Y (MgCO3) = 294 / 84.3 = 3.48 mol;
We compose the proportion based on the data on the reaction equation:
3.48 mol (MgCO3) – X mol (CO2);
Hence, X mol (CO2) = 3.48 * 1/1 = 3.48 mol
Find the volume of CO2: V (CO2) = 3.48 * 22.4 = 77.95 l
Answer: when MgCO3 is ignited, CO2 is released with a volume of 77.95 liters.