Chapter 20.7, Problem 2.1ACP

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Assume that a sample of hard water contains 50. mg/L of Mg2+ and 150 mg/L of Ca2+, with HCO3− as the accompanying anion. What mass of CaO should be added to 1.0 L of this aqueous solution to precipitate all the Mg2+ and Ca2+ as CaCO3 and MgCO3? What is the total mass of the two solids formed?

Interpretation Introduction

Interpretation:

The mass of CaO should be added to 1L of the given solution to precipitate all Mg2+andCa2+ and the total mass of two solid formed in the given reaction should be determined

Concept introduction:

• From its given mass is,

Number of moles=GivenmassMolecularmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
Explanation

Given,

A sample of hard water contains 50.â€‰mg/Lâ€‰ofâ€‰â€‰Mg2+â€‰andâ€‰â€‰150â€‰â€‰mg/lâ€‰â€‰â€‰ofâ€‰â€‰Ca2+,â€‰â€‰â€‰withâ€‰â€‰HCO3âˆ’Â  as accompanying anion.

Balanced chemical equation for the reaction involved in between Mg2+â€‰â€‰andâ€‰â€‰â€‰Ca2+ with HCO3âˆ’

â€‚Â Ca(aq)2+â€‰+â€‰2HCO3(aq)-+â€‰CaO(s)â€‰â€‰â€‰â†’â€‰2CaCO3(s)â€‰+H2O(l)Mg(aq)2+â€‰+â€‰2HCO3(aq)-+â€‰CaO(s)â€‰â€‰â†’2CaCO3(s)â€‰+â€‰MgCO3(s)+H2O(l)

The amount (in mg) of CaO available for Mg2+â€‰â€‰andâ€‰â€‰â€‰Ca2+ can be determined as follows,

â€‚Â Numberâ€‰ofâ€‰moleâ€‰=Givenâ€‰massâ€‰ofâ€‰theâ€‰substanceMolarâ€‰massâ€‰

Â Â Forâ€‰â€‰Mg2+â€‰â€‰:â€‰â€‰50â€‰mgâ€‰Ã—â€‰1â€‰mmolâ€‰â€‰Mg2+24.31â€‰â€‰mgâ€‰Ã—â€‰1â€‰mmolâ€‰â€‰CaOmmolâ€‰â€‰Mg2+â€‰Ã—â€‰56.08â€‰â€‰mg/mmolâ€‰CaO â€‰â€‰â€‰â€‰â€‰â€‰â€‰:â€‰â€‰â€‰115â€‰â€‰mgâ€‰â€‰CaOForâ€‰â€‰Ca2+â€‰â€‰:â€‰â€‰150â€‰mgâ€‰Ã—â€‰1â€‰mmolâ€‰â€‰Ca2+40.08â€‰â€‰mgâ€‰Ã—â€‰1â€‰mmolâ€‰â€‰CaOmmolâ€‰â€‰Ca2+â€‰Ã—â€‰56.08â€‰â€‰mg/mmolâ€‰CaO â€‰â€‰â€‰â€‰â€‰â€‰â€‰:â€‰â€‰â€‰210â€‰mgâ€‰â€‰CaOTotalâ€‰â€‰massâ€‰â€‰ofâ€‰â€‰CaOâ€‰â€‰=â€‰â€‰115â€‰mgâ€‰+â€‰210â€‰mgâ€‰ â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰â€‰325â€‰â€‰mg

325â€‰â€‰mg of CaO should be added to 1â€‰L of the given solution to precipitate all â€‰Mg2+â€‰andâ€‰Ca2+ as MgCO3â€‰â€‰andâ€‰â€‰CaCO3

Similar to the above calculation the mass of solids produced (precipitated) in the given reaction can be calculated as follows,

From the balanced equation of the given reaction, it is clear that 2â€‰â€‰molâ€‰â€‰CaCO3â€‰ per mole â€‰Ca2+ and 1â€‰â€‰molâ€‰eachâ€‰â€‰ofâ€‰â€‰CaCO3â€‰â€‰andâ€‰â€‰â€‰MgCO3â€‰â€‰perâ€‰â€‰moleâ€‰â€‰Mg2+

CaCO3â€‰â€‰â€‰fromâ€‰â€‰Ca2+â€‰â€‰reactionâ€‰â€‰is,(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