The weight of MgCO₃ required for the preparation of 12g of MgSO₄ using H₂SO₄ is

$$\begin{gathered} {\mathrm{MgCO}}_3 +{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{MgSO}}_4+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2 \\ 1 \mathrm{mole} \mathrm{of} {\mathrm{MgCO}}_3 -3 \mathrm{moles} \mathrm{of} {\mathrm{MgSO}}_4 \\ 84 \mathrm{gms} \mathrm{of} {\mathrm{MgCO}}_3-120 \mathrm{gms} \mathrm{of} {\mathrm{MgSO}}_4 \\ \mathrm{X} \mathrm{gms} \mathrm{of} {\mathrm{MgCO}}_3 -12 \mathrm{gms} \mathrm{of} {\mathrm{MgSO}}_4 \\ \mathrm{X}=\frac{12}{120}\times 84 =8.4 \mathrm{gms} \end{gathered}$$

X = 8.4g