100.0 cm3 of a 0.50 mol dm-3 solution of BaCl2 is added to a 50.0 cm3 of a 0.10 mol dm-3 solution of Na2SO4. A precipitate of BaSO4 is formed according to the equation below BaCl2(aq) + Na2SO4(aq) --> BaSO4(s) + 2NaCl(aq) What amount, in mol, of BaSO4 is produced?

100.0 cm3 of a 0.50 mol dm-3 solution of BaCl2 is added to a 50.0 cm3 of a 0.10 mol dm-3 solution of Na2SO4. A precipitate of BaSO4 is formed according to the equation below BaCl2(aq) + Na2SO4(aq) --> BaSO4(s) + 2NaCl(aq) What amount, in mol, of BaSO4 is produced?

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 94AP

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Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

100.0 cm³ of a 0.50 mol dm^-3solution of BaCl₂ is added to a 50.0 cm³ of a 0.10 mol dm^-3solution of Na₂SO₄. A precipitate of BaSO₄is formed according to the equation below

BaCl_2(aq)+ Na₂SO_4(aq)--> BaSO_4(s)+ 2NaCl_(aq)

What amount, in mol, of BaSO₄is produced?

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