   Chapter 9, Problem 48QAP ### 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
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# For each of the following unbalanced chemical equations, suppose that exactly 1.00 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed).3msp;  CS 2 ( l ) + O 2 ( g ) → CO 2 ( g ) + SO 2 ( g ) msp;  NH 3 ( g ) + CO 2 ( g ) → CN 2 H 4 O ( s ) + H 2 O ( g ) msp;  H 2 ( g ) + MnO 2 ( s ) → MnO ( s ) + H 2 O ( g ) msp;  I 2 ( l ) + Cl 2 ( g ) → ICI ( g )

Interpretation Introduction

(a)

Interpretation:

The limiting reactant and the amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Explanation

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

CS(l)+ 3O2  (g)CO2 (g)+ 2SO2(g)

Given:

Amount of CS = 1.00 g

Amount of O2 = 1.00 g

Calculation:

Number of moles of CS and O2 calculated as follows:

Number of moles=mass in gmolarmass=1.00 g76.139 g/mol=0.0131 moles CS2 Number of moles=mass in gmolarmass=1

Interpretation Introduction

(b)

Interpretation:

The limiting reactant and the amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant and the amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS + 3 O22PbO + 2SO2 Reactants Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant and the amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

2 PbS +  3 O22PbO +  2SO2  Reactants            Products

Mass of any substance can be calculated as follows:

Mass in gram = Number of moles×Molar mass

Number of moles can be calculated as follows;

Number of moles=mass in gmolarmass.

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