   # What mass of barium sulfate can be produced when 100.0 mL of a 0.100 -M solution of barium chloride is mixed with 100.0 mL of a 0.100- M solution of iron(III) sulfate? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 6, Problem 60E
Textbook Problem
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## What mass of barium sulfate can be produced when 100.0 mL of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?

Interpretation Introduction

Interpretation: To find out the mass of barium sulphate can be produced when 100ml of a 0.100M solution of barium chloride is added to 100ml  of a 0.100M solution of iron (III) sulfate.

Concept Introduction:  The precipitate will be formed when two soluble salt solutions combined together is called as a precipitation reaction.

Formula:

Mole=volumeofsolution (L)×molarity (M)litre (L) (1)

Mass = required number of mole ×molecularweightrequired moleratio (2)

### Explanation of Solution

Explanation

Volume = 0.100L

Molarity = 0.100M

Substitute volume and molarity in formula (1) we get the mole of solutions.

Mole=volumeofsolution (L)×molarity (M)litre (L) (1)

Volume of BaCl2 solution is multiply with its concentration then given values is divided by 1 we get mole of BaCl2 solution

0.100(L)×0.100(M)1(L)

Substitute the volume and molarity of solution in mole formula (1) to get mole of BaCl2 solution

Mole of BaCl2 solution = 0.01mole

Calculate the mole of iron (III) sulphate solution as follows,

Volume = 0.100L

Molarity = 0.100M

Substitute the volume and molarity of solution in mole formula (1) to get mole of iron

(III) sulphate solution

Volume of iron (III) sulphate solution is multiply with its concentration then given values is divided by 1 we get mole of iron (III) sulphate solution

= 0

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