Question

the volume of carbon dioxide at ntp involved by strong heating of 20 grams of calcium carbonate is​

Answer 1

The volume of carbon dioxide at ntp evolved by strong heating of 20 grams of calcium carbonate is 4.48 L.

• Given weight of calcium carbonate heated = 20 g

• The balanced equation for the heating of calcium carbonate is :

CaCO3 + ∆  →  CaO + CO2

• From the equation, we see that one mole of calcium carbonate evolves one mole of carbon dioxide.

• We know that one mole of CaCO3 weighs 100 g.

• Therefore, number of moles present in 20 g of CaCO3 = 20 g / 100 g = 0.2 moles

=> Number of moles of CO2 given by 0.2 moles of CaCO3 = 0.2 moles

• According to Avogadro's law for gases, one mole of any gas at normal temperature and pressure occupies 22.4 L of volume.

• Therefore, volume occupied by 0.2 moles of CO2 = 0.2 × 22.4 L = 4.48 L

Answer 2

answer is 4.48 L

hope it was helpful