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The reaction between the red lead and hydrochloric acid is given below: PbA3OA4+8HCl⟶3PbClA2+4HA2O+ClA2 Calculate the mass of lead chloride formed by the action of the 6.85 g of red lead.

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Question

The reaction between the red lead and hydrochloric acid is given below:

\[\ce{Pb3O4 + 8HCl -> 3PbCl2 + 4H2O + Cl2}\]

Calculate the mass of lead chloride formed by the action of the 6.85 g of red lead.

Numerical
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Solution

Molecular mass of Pb3O4 = 3 207.2 + 4 16 = 685 g

685 g of Pb3O4 gives = 834 g of PbCl2

Hence, 6.85 g of Pb3O4 will give = `(6.85 xx 834)/685` = 8.34 g

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Chapter 5: Mole concept and Stoichiometry - EXERCISE-5D [Page 98]

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S.P. Singh Concise Chemistry [English] Class 10 ICSE
Chapter 5 Mole concept and Stoichiometry
EXERCISE-5D | Q 3. (a) | Page 98
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