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MnOA2+4HCl⟶MnClA2+2HA2O+ClA2 0.02 moles of pure MnO2 is heated strongly with conc. HCl. Calculate moles of chlorine gas formed.

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Question

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

0.02 moles of pure MnO2 is heated strongly with conc. HCl. Calculate moles of chlorine gas formed.

Numerical
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Solution

1 mole Mn02 used, Cl2 formed = 1 mole

∴ 0.02 mole MnO2 will form = 0.02 mole of Cl2

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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. (d) | Page 94

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