   Chapter 9, Problem 49QAP ### 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
8 views

# For each of the following unbalanced chemical equations, suppose 1.00 g of each reactant is taken. Show by calculation which reactant is limiting. Calculate the mass of each product that is expected.l type='a'>   UO 2 ( s ) + HF ( a q ) → UF 4 ( a q ) + H 2 O ( l ) i>   NaNO 3 ( a q ) + H 2 SO 4 ( a q ) → Na 2 SO 4 ( a q ) + HNO 3 ( a q ) i>   Zn ( s ) + HCl ( a q ) → ZnCl 2 ( a q ) + H 2 ( g ) i>   B ( OH ) 3 ( s ) + CH 3 OH ( l ) → B ( OCH 3 ) 3 ( s ) + H 2 O ( l )

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 reaction has due to following properties:

1. Limiting reactant completely reacted in a 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:

UO2(s) + 4 HF (l) UF4(s) + 2 H2O(l)

Given:

Amount of UO2 = 1.00 g

Amount of HF = 1.00 g

Calculation:

Number of moles of UO2 and HF calculated as follows:

Number of moles=mass in gmolarmass=1.00 g253.8089 g/mol=0.00394 moles UO2Number of moles=mass in gmolarmass=1.00 g20.01 g/mol=0.0499 moles HF

In this reaction UO2 is a limiting reactant because it completely reacted in the reaction

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

(d)

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.

### 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 