For reaction MnO2+4HCl⟶MnCl2+2H2O+Cl2(g)261 gm MnO2 is mixed with 448 litres of HCl gas at 273∘C and 1 atm pressure to produce product [NA= 6×1023; Atomic mass of Mn=55,Cl=35.5]Select correct statement(s).

MnO 2 + 4 HCl → MnCl 2 + 2 H 2 O + Cl 2 ( g ) n = 261 87 224 22.4 = 3 = 10 0.5 mole (L.R) 10 4 × N A Hence, options B and C are correct.

Chlorine gas produced contains 15 × 10 23 molecules.

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For reaction ${MnO}_{2} + 4 HCl ⟶ {MnCl}_{2} + 2 H_{2} O + {Cl}_{2} (g)$
$261 gm {MnO}_{2}$ is mixed with 448 litres of $HCl$ gas at $273^{\circ} C$ and $1 atm$ pressure to produce product $[N_{A} = 6 \times 10^{23};$ Atomic mass of $Mn = 55, Cl = 35.5]$
Select correct statement(s).

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Moles of excess reactant left is $0.5$ moles.

If $%$ yield of reaction is $50 %$ , then mass of $M n C l_{2}$ obtained will be 315.

Chlorine gas produced contains $15 \times 10^{23}$ molecules.

$M n O_{2}$ is limiting reagent.

answer is B, C.

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${MnO}_{2} + 4 HCl \to {MnCl}_{2} + 2 H_{2} O + {Cl}_{2} (g)$

$\begin{aligned} n = \frac{261}{87} \frac{224}{22.4} \\ = 3 = 10 \end{aligned}$

$0.5 mole (L.R) \frac{10}{4} \times N_{A}$

Hence, options B and C are correct.

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