Billions of pounds of urea, CO ( NH 2 ) 2 , are produced annually for use as a fertilizer. The principal reaction employed is: 2NH 3 + CO 2 → CO ( NH 2 ) 2 + H 2 O By assuming unlimited amounts of CO 2 , how many moles of urea can be produced from each of the following amounts of NH 3 ? a . 2 mol NH 3 b . 0 .45 mol NH 3 c . 10 g NH 3 d . 2 .0 kg NH 3

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Answer 1

Textbook Question

Chapter 4, Problem 54E

Billions of pounds of urea, $CO ( NH 2 ) 2 ,$ are produced annually for use as a fertilizer. The principal reaction employed is:

$2NH 3 + CO 2 → CO ( NH 2 ) 2 + H 2 O$

By assuming unlimited amounts of CO₂, how many moles of urea can be produced from each of the following amounts of $NH 3$ ?

$a . 2 mol NH 3 b . 0 .45 mol NH 3 c . 10 g NH 3 d . 2 .0 kg NH 3$

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

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Answer 2

Textbook Question

Chapter 4, Problem 54E

Billions of pounds of urea, $CO ( NH 2 ) 2 ,$ are produced annually for use as a fertilizer. The principal reaction employed is:

$2NH 3 + CO 2 → CO ( NH 2 ) 2 + H 2 O$

By assuming unlimited amounts of CO₂, how many moles of urea can be produced from each of the following amounts of $NH 3$ ?

$a . 2 mol NH 3 b . 0 .45 mol NH 3 c . 10 g NH 3 d . 2 .0 kg NH 3$

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

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Answer 3

Textbook Question

Chapter 4, Problem 54E

Billions of pounds of urea, $CO ( NH 2 ) 2 ,$ are produced annually for use as a fertilizer. The principal reaction employed is:

$2NH 3 + CO 2 → CO ( NH 2 ) 2 + H 2 O$

By assuming unlimited amounts of CO₂, how many moles of urea can be produced from each of the following amounts of $NH 3$ ?

$a . 2 mol NH 3 b . 0 .45 mol NH 3 c . 10 g NH 3 d . 2 .0 kg NH 3$

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

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Answer 4

Textbook Question

Chapter 4, Problem 54E

Billions of pounds of urea, $CO ( NH 2 ) 2 ,$ are produced annually for use as a fertilizer. The principal reaction employed is:

$2NH 3 + CO 2 → CO ( NH 2 ) 2 + H 2 O$

By assuming unlimited amounts of CO₂, how many moles of urea can be produced from each of the following amounts of $NH 3$ ?

$a . 2 mol NH 3 b . 0 .45 mol NH 3 c . 10 g NH 3 d . 2 .0 kg NH 3$

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The number of moles of urea is to be determined in the given amounts of $NH3$ .

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer 12

Answer to Problem 54E

Solution: The number of moles of urea present in the molecules is given as follows:

$1 mol urea$

$0.23 mol urea$

$0.3 mol urea$

$59 mol urea$

Answer 13

Explanation of Solution

a) $2 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 mol NH3×1 mol urea2 mol NH3=1 mol urea$

Therefore, the number of moles of urea is $1 mol$ .

b) $0.45 mol NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$0.45 mol NH3×1 mol urea2 mol NH3=0.23 mol urea$

Therefore, the moles of urea are $0.23 mol$ .

c) $10 g NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$10 g NH3×1 mol NH317.03 g NH3×1 mol urea2 mol NH3=0.3 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $0.3 mol$ .

d) $2.0 kg NH3$

The equivalence is obtained between grams of $NH3$ and moles of $NH3$ with the help of the chemical formula as follows:

$1 mol NH3≡17.03 g NH3$

The equivalence is obtained between moles of $NH3$ and moles of urea with the help of the chemical formula as follows:

$1 mol urea≡2 mol NH3$

In order to obtain the moles of urea, the conversion factor proceeds as follows:

$2 kg NH3×(1000 g1 kg)×(1 mol NH317.03 g NH3)×(1 mol urea2 mol NH3)=59 mol urea$

Here,

$molar mass of NH3=17.03 g/mol$

Therefore, the moles of urea are $59 mol$ .

Answer 14

Ch. 4 - Interpreting Chemical Formulas Determine the...Ch. 4 - Prob. 4.2YT Ch. 4 - Prob. 4.3YT Ch. 4 - Prob. 4.4YT Ch. 4 - Calculating Formula Mass Calculate the formula...Ch. 4 - Using the Molar Mass to Find the Number of...Ch. 4 - Prob. 4.7YT Ch. 4 - Chemical Formulas as Conversion Factors (Mass to...Ch. 4 - Prob. 4.9YT Ch. 4 - Prob. 4.10YT

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Answer to Problem 54E

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