If 0.80 mole of MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O then

MnO 2 + 4HCl → MnCl 2 + Cl 2 + 2H 2 O ∴ MnO 2 is limiting reagent

If 0.80 mole of $M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$ then

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0.80 mole of $C l_{2}$ is formed

0.80 mole of HCl remains unreacted

$M n O_{2}$ is completely reacts

$M n O_{2}$ is the limiting reactant

answer is A, B, C, D.

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

$∴ {MnO}_{2}$ is limiting reagent

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