If 0.80 mole of ${\mathrm{MnO}}_2$ and 146 g of HCl react

${\mathrm{MnO}}_2+4\mathrm{HCl}\to {\mathrm{MnCl}}_2+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}$

${\mathrm{MnO}}_2+4\mathrm{HCl}\to {\mathrm{MnCl}}_2+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}$

1 mole      4 mole 

0.8 mole   3.2 mole

0.8 mol ${\mathrm{MnO}}_2=3.2 \mathrm{mole} \mathrm{HCl}=3.2\times 36.5=1 \mathrm{mole} {\mathrm{Cl}}_2$

=116.8 grams

Given ${\mathrm{n}}_{\mathrm{HCl}}=\frac{146}{36.5}=4 \mathrm{moles}$

The limiting agent = ${\mathrm{MnO}}_2$

Excess of HCl remains = 146 − 116.8

= 29.2 grams

(or) $\frac{29.2}{36.5}=0.8 moles of HCl remains$