   Chapter 15.6, Problem 15.8SC ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
23 views

# Exercise 15.8 When aqueous solutions of Na2SO3 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M NaSO4 are mixed.

Interpretation Introduction

Interpretation:

The mass of PbSO4 which is formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed should be calculated.

Concept Introduction:

Molarity of the solution is equal to the ratio of number of moles of solute to the volume of solution in litres.

The mathematical expression is given by:

Molarity = number of moles of solute volume of solution in litres

Number of moles = given massmolar mass.

Explanation

Given information:

Molarity of Pb(NO3)2 = 0.0500 M

Volume of Pb(NO3)2 = 1.25 L

Molarity of Na2SO4 = 0.0250 M

Volume of Na2SO4 = 2.00 L

When aqueous solution of both Na2SO4 having Na+ and SO42 ions and, Pb(NO3)2 having Pb2+ and NO3 are mixed together, PbSO4 is formed.

The reaction is shown below:

Pb2+(aq)+SO42(aq)PbSO4(aq)

To identify the limiting reagents, first determined the number of moles of Pb2+ and SO42

Since, 0.0500 M of Pb(NO3)2 consist of 0.0500 M Pb2+ in 1.25 L.

Molarity = number of moles of solute volume of solution in litres

Number of moles of Pb2+ ion = molarity×volume of solution in litres

=0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 