the volume of carbon dioxide at ntp involved by strong heating of 20 grams of calcium carbonate is
Answers
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 is 4.48 L
hope it was helpful
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