Mass of $HCl$ in $500$ cc of $0.5 \,N \,HCI $
$w=\frac{0.5 \times 36.5 \times 500}{1000}\left[\because \quad N=\frac{w \times 1000}{e q . w t . x V(\text { incc })}\right]$
$ = 9.125 \, g$
The balanced equation for the reaction is
$\underset{100 g }{ Ca } CO _3+\underset{2 \times 36.5 g }{2 HCl } \longrightarrow CaCl _2+ H _2 O + CO _2$
${2 \times 36.5\, g}$ of $HCI$ react with $100 \,g $ of $CaCO_3$
$ \therefore 9.125\, g$ of $HCI $ will react with
$ \frac{100 \times 9.125}{2 \times 36.5} = 12.5 \, g$ to $CaCO_3$