Chapter 22.7, Problem 3.3ACP

### 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

# The compound (NH4)2Ce(NO3)6 is a widely used oxidizing agent in redox titrations. A 0.181 g sample of pure iron is dissolved in acid, reduced to Fe2+, and then titrated with 31.33 mL of (NH4)2Ce(NO3)6 (to give Ce3+ and Fe3+). Calculate the concentration of the cerium salt.

Interpretation Introduction

Interpretation:

Concentration cerium salt in the in the given compound has to be calculated..

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

Molarity=MassperlitreMolecular mass

• A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,

Numberofmole=GivenmassofthesubstanceMolarmass

• 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).
• Concentrationofsubstance=Amountof substancevolumeofthesubstance
• 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.
Explanation

Probable equation for the given reaction can be written as follows,

Fe(aq)2+â€‰+â€‰[Ce(NO3)6](aq)2âˆ’â€‰â†’â€‰Fe(aq)3++â€‰Ce(aq)3+â€‰+â€‰6NO3(aq)âˆ’

Given mass of Feâ€‰â€‰isâ€‰â€‰0.181â€‰g which is dissolved in acid, reduced to Fe2+ and then titrated with 31.33â€‰â€‰mLâ€‰â€‰ofâ€‰â€‰(NH4)2Ce(NO3)6

Amount of Fe2+ (in mol) can be calculated as follows,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  0.181â€‰gâ€‰Feâ€‰Ã—1â€‰â€‰molâ€‰ofâ€‰â€‰Feâ€‰55.85â€‰gâ€‰Feâ€‰=â€‰â€‰0

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