Question

3.68 g of a mixture of caco3 and mgco3 is heated to librate 0.04 mole of co2 .the mole % of caco3 and mgco3 in the mixture is respectively​

Answer 1

Answer:

Explanation:

We know that

Molar mass of Caco3 (calcium carbonate) = 100 g/mol

Molar mass of Mgco3(magnesium carbonate) = 84.3 g/mol

Now according to question

2 moles of carbon dioxide CO2 requires 1 mole of calcium carbonate

So 0.04 mole of CO2 requires 0.04 / 2 = 0.02 moles of CaCO3

Similarly 0.04 mole of CO2 requires 0.02 mole of MgCO3

Mass of a substance = molar mass x number of moles

So mass of CaCO3 used = 100 g/mol x 0.02 mol = 2 g

Also mass of MgCO3 used = 84.3 g/mol x 0.02 mol = 1.686 g

Now mass of reacting mixture = 2 + 1.686 = 3.68 g  

So it means 0.02 moles each of MgCO3 and CaCO3 are present in the mixture.

So it will be 50% 50%