Question

What volume of carbon dioxide is produced when 37.5 g of calcium carbonate reacts completely according to the following reaction at 25 °C and 1 atm?

calcium carbonate (s) calcium oxide (s) + carbon dioxide(g)

liters carbon dioxide

Answer 1

CaCO3(s) -> CaO(s) + CO2(g)

Moles of CaCO3 = mass/molar mass

= 37.5/100 = 0.375 moles

According to the reaction,

1 mole CaCO3 give 1 mole CO2

0.375 mole CaCO3 give 0.375 moles of CO2

Now, Using ideal gas equation :

Volume, V = nRT/P

Where , n = 0.375, R = 0.082 Latm/mol.K, P = 1 atm

T = 25°C = 25+273.15 K = 298.15 K

So, Volume, V = 0.375×0.082×298.15/1

= 9.17 L