Find the weight of 75%pure magnesium carbonate which generates 50g magnesium oxide on heating

Answer 1

Magnesium carbonate = MgCO₃ (MW = 84.3 g/mol)

Magnesium oxide = MgO (MW = 40.3 g/mole)

Reaction: MgCO₃ ==> MgO + CO₂

mols MgO produced = 50 g MgO x 1 mol MgO / 40.3 g = 1.24 mols MgO produced

mols MgCO3 needed to produce this = 1.24 mols MgO x 1 mol MgCO3 / mol MgO = 1.24 mols MgCO3

Mass MgCO3 needed (if pure) = 1.24 mols MgCO3 x 84.3 g / mol = 104.5 g pure MgCO3

Since the MgCO3 is only 75% pure, we need to correct for this as follows:

104.5 g / 0.75 = 139 g (this is to 3 sig.figs., but 50 g MgO has only 1 sig. fig. so answer should be 100 g)