Weight of 60% pure Magnesite  required to neutralize two moles of HCl will be

$$\begin{gathered} {\mathrm{MgCO}}_3+2\mathrm{HCl} \to {\mathrm{MgCl}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2; \\ 84 \mathrm{g} 2 \mathrm{moles} \end{gathered}$$

but the purity is only 60%, more amount of magnesite is required,

weight of magnesite required = $84\times \frac{100}{60}=140 \mathrm{gr}$