Question

What weight of sulphuric acid will be required to dissolve 3 grams of magnesium carbonate?
[Mg = 24, C = 12, O = 16]
MgCO₃ + H₂SO₄ $\stackrel{}{\to }$ MgSO₄ + H₂O + CO₂

Answer 1

Number of moles in 3 g of magnesium carbonate = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{magnesium}\mathrm{carbonate}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{magnesium}\mathrm{carbonate}}=\frac{3}{24+12+\left(3\times 16\right)}=0.036$

According to the balanced chemical equation, one mole of sulphuric acid is required to dissolve one mole of magnesium carbonate.
Number of moles of sulphuric acid needed to dissolve 3 g of magnesium carbonate = 0.036
Mass of sulphuric acid in 0.036 moles = Number of moles $\times$ Molecular mass of sulphuric acid = (0.036 $\times$ 98) g = 3.5 g