100.0 g of a ground mineral sample consisting of a mixture of MgCO₃ (magnesite) and BaCO₃ (witherite) was heated to completely convert the carbonates to their metal oxides

Question

After the reaction 60.95 g of the metal oxide mixture was obtained. What was the mass of MgCO₃ in the original sample?

Answer 1

MgCO₃ ==> MgO + CO₂

BaCO₃ ==> BaO + CO₂

molar mass MgCO₃ = 84.3 g/mol

molar mass BaCO₃ = 197 g/mol

molar mass MgO = 40.3 g/mol

molar mass BaO = 153 g/mol

let xg = g BaCO₃

100-x = g MgCO₃

x g BaCO₃/197 gmol^-1 = mols BaCO₃ = mols Ba = mols BaO

100-x g MgCO₃/84.3 gmol^-1 = mols MgCO₃ = mols Mg = mols MgO

x g / 197 gmol^-1(153 gmol^-1) + 100-x g / 84.3 gmol^-1)(40.3 gmol^-1) = 60.95 g

x = 43.98 g BaCO₃

100-x = 56.02 g MgCO₃

I believe this is the correct approach to this problem, but I strongly urge you to check my math!!!