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ChemistryIntroductory Chemistry: A Foundation9th Edition

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Chapter 9, Problem 52QAP
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Introductory Chemistry: A Foundati...

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

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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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Copper(II) sulfate has been used extensively as a fungicide (kills fungus) and herbicide (kills plants). Copper(II) sulfate can be prepared in the laboratory by reaction of copper(II) oxide with sulfuric acid. The unbalanced equation is

:math> C u O ( s ) + H 2 SO 4 ( a q ) → CuSO 4 ( a q ) + H 2 O ( l )

2.49 g of copper(II) oxide is treated with 5.05 g of pure sulfuric acid, which reactant would limit the quantity of copper(II) sulfate that could be produced?

Interpretation Introduction

Interpretation:

The reactant that limits the quantity of copper (II) sulphate produced should be determined.

Concept Introduction:

Limiting reactant is the reactant that limits the amount of product formed in the chemical reaction. The mass of product formed from limiting reactant is theoretical yield which is less than the mass of product formed from the reactant in excess.

Explanation

The balanced chemical reaction is as follows:

CuO(s)+H2SO4(aq)→CuSO4(aq)+H2O(l)

The mass of copper (II) oxide is 2.49 g and that of mass of pure sulphuric acid is 5.05 g. To determine the limiting reactant, first calculate the number of moles of copper (II) oxide and sulphuric acid as follows:

Molar mass of copper (II) oxide is 79.545 g/mol thus, number of moles will be:

n=mM

Putting the values,

n=2.49 g79.545 g/mol=0.0313 mol

From the balanced chemical reaction, 1 mol of CuO gives 1 mol of CuSO4 thus, 0.0313 mol of CuO produces 0.0313 mol of CuSO4.

Since, molar mass of CuSO4 is 159.61 g/mol, mass of CuSO4 will be:

m=n×M

Putting the values,

m=(0.0313 mol)(159

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