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MnOA2+4HCl⟶MnClA2+2HA2O+ClA2 0.02 moles of pure MnO2 is heated strongly with conc. HCl. Calculate the mass of acid required.

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Question

\[\ce{MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2}\]

0.02 moles of pure MnO2 is heated strongly with conc. HCl. Calculate the mass of acid required.

Numerical
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Solution

For 1 mole MnO, acid required = 4 mole of HCl

So, for 0.02 mole, acid required = 4 × 0.02 = 0.08 mole

Mass of HCl = 0.08 x 36.5 = 2.92 g

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Chapter 5: Mole concept and Stoichiometry - EXERCISE-5D [Page 94]

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S.P. Singh Concise Chemistry [English] Class 10 ICSE
Chapter 5 Mole concept and Stoichiometry
EXERCISE-5D | Q 11. (h) | Page 94

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