Question

1.84 gram equimolar mixture of CaCO3 and MgCO3 on heading gives CO2.Volume of CO2 Obtained is measured to be 448ml at STP.Mass of CaCO3 in mixture is-

Answer 1

Given:

• Total mass of the mixture (m) = 1.84g
• Moles of MgCO₃ = Moles of CaCO₃ = n
• Volume of CO₂ obtained at STP (v) = 448 mL
• Molar mass of MgCO₃ (m1) = 84 g/mol
• Molar mass of CaCO₃ (m2) = 100 g/mol

To find:

Mass of CaCO₃ in the mixture.

Solution:

• Each mole of MgCO₃ as well as CaCO₃ produces one mole of CO₂ gas on heating.
• Moles of CO₂ produced = 2n
• 22.4 L of volume is occupied by one mole of any gas at STP. So, moles of CO₂ = v (in L)/22.4 = 0.02
• 0.02 = 2n ⇒ n = 0.01
• Mass of CaCO₃ = m2*n = 100*0.01 = 1 g

Answer:

Mass of CaCO₃ in the mixture is equal to 1 g.