   Chapter 15, Problem 67QAP ### 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
294 views

# 67. When aqueous solutions of lead(II) ion are treated with potassium chromate solution, a bright yellow precipitate of lead(II) chromate, PbCrO4, forms. How many grains of lead chromate form when a 1.00-g sample of Pb(NO3)2 is added to 25.0 mL of 1 .00 M K2CrO4 solution?

Interpretation Introduction

Interpretation:

The grams of lead chromate produced when 1.00-g sample of PbNO32 is added to 25.0mL of 1.00MK2CrO4 in the given solution are to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolarmass

Molarity is used to find out the concentration of solution.

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL.

Explanation

The molarity and volume of K2CrO4 solution is given to be 1.00M and 25.0mL respectively.

The conversion of units of 25.0mL into L is done as,

25.0mL=25.01000L=0.025L

The molar mass of PbNO32 is 331.2g/mol.

The moles of lead nitrate PbNO32 is calculated by the formula,

Moles=MassgMolarmass        (1)

Substitute the values of mass and molar mass in the equation (1)

Moles=1.00g331.2g/mol=0.00302moles

The number of moles of K2CrO4 solution is calculated by the formula,

Numberofmoles=Molarity×Volumeofsolution        (2)

Substitute the values of molarity and volume of K2CrO4 solution in the equation (2).

Numberofmoles=1.00M×0.025L=0.025moles

The chemical equation of reaction between lead nitrate and potassium chromate is shown below

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