The weight of MgCO₃ required for the preparation of 1.2 gm of MgSO₄ is

$MgC{O}_3+H_{2}S{O}_4\underset{}{\overset{}{\to }}MgS{O}_4+C{O}_2+H_{2}O$

               120 gm MgSO₄ obtained from $\underset{}{\overset{}{\to }}$  84 gm MgCO₃

               1.20 gm MgSO₄ obtained from $\underset{}{\overset{}{\to }}$ ?

                                                                   $=\frac{84}{120}\times 1.2=0.84$ gm