How much carbon monoxide (4) can be obtained from the decomposition of 160 g of calcium carbonate (limestone)?

Given:
m (CaCO3) = 160 g
Vm = 22.4 l / mol
Find: V (CO2) -?
Decision:
1) Write the equation of a chemical reaction:
CaCO3 => CaO + CO2 ↑;
2) Calculate the molar mass of CaCO3:
M (CaCO3) = Mr (CaCO3) = Ar (Ca) + Ar (C) + Ar (O) * 3 = 40 + 12 + 16 * 3 = 100 g / mol;
3) Calculate the amount of CaCO3 substance:
n (CaCO3) = m (CaCO3) / M (CaCO3) = 160/100 = 1.6 mol;
4) Determine the amount of substance CO2 (taking into account the coefficients in the reaction equation):
n (CO2) = n (CaCO3) = 1.6 mol;
5) Calculate the volume of CO2 (under normal conditions):
V (CO2) = n (CO2) * Vm = 1.6 * 22.4 = 35.84 liters.
Answer: The CO2 volume is 35.84 liters.