Calculate the volume of CO2 produced at STP from the decomposition of 152 g calcium carbonate caco3?

Answer 1

Molar mass CaCO3 = 100.087 g/mol

Moles CaCO3 = 152 g / 100.087 = 1.52

the ratio between CaCO3 and CO2 is 1 : 1 so we get 1.52 moles of CO2

At STP p=1 ATM and T = 273 K

V = nRT / p = 1.52 x 0.0821 x 273 /1 = 34.1L

Answer 2

Using the balanced chemical equation for the decomposition of calcium carbonate, 1 mole of CaCO3 produces 1 mole of CO2. The molar mass of CaCO3 is 100 g/mol, therefore 152 g of CaCO3 is 1.52 moles. At STP, 1 mole of any gas occupies 22.4 L, so 1.52 moles of CO2 will occupy 34.05 L.