Chapter 9, Problem 15CR

Interpretation Introduction

(a)

Interpretation:

The number of moles in 2.45 g ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{a}\text{m}\text{u}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0153{\text{ mole of Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

Calculation:

Number of moles can be calculated as follows;

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}\\ =\frac{\text{2}\text{.45 g}}{\text{159}\text{.69 g/mol}}\\ =0.0153{\text{ mole of Fe}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

The number of moles in 2.45 g of ${\text{P}}_4(s)$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{ amu}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0198{\text{ mole of P}}_{\text{4}}(s)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g ${\text{P}}_{\text{4}}(s)$

Calculation:

Number of moles can be calculated as follows:

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}\\ =\frac{\text{2}\text{.45 g}}{\text{123}\text{.90 g/mol}}\\ =0.0198{\text{ mole of P}}_{\text{4}}(s)\end{array}$.

Interpretation Introduction

(c)

Interpretation:

The number of moles in 2.45 g of ${\text{Cl}}_{\text{2}}(g)$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{ amu}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0346\text{ mole of }\text{C}{\text{l}}_2(g)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g $\text{C}{\text{l}}_2(g)$

Calculation:

Number of moles can be calculated as follows;

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}\\ =\frac{\text{2}\text{.45 g}}{\text{70}\text{.906 g/mol}}\\ =0.0346{\text{ mole of Cl}}_{\text{2}}(g)\end{array}$.

Interpretation Introduction

(d)

Interpretation:

The number of moles in 2.45 g of ${\text{Hg}}_{\text{2}}\text{O(}\text{s})$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{ amu}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0587{\text{ mole of Hg}}_{\text{2}}\text{O}(s)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g $\text{H}{\text{g}}_2\text{O}(s)$

Calculation:

Number of moles can be calculated as follows;

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}\\ =\frac{\text{2}\text{.45 g}}{\text{417}\text{.183 g/mol}}\\ =0.0587{\text{ mole of Hg}}_{\text{2}}\text{O}(s)\end{array}$.

Interpretation Introduction

(e)

Interpretation:

The number of moles in 2.45 g of $\text{HgO}$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{ amu}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0113\text{ mole of HgO}(s)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g $\text{HgO}$

Number of moles can be calculated as follows;

.

Interpretation Introduction

(f)

Interpretation:

The number of moles in 2.45 g of ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(s)$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.00\text{8 amu}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0149{\text{ mole of Ca(NO}}_{\text{3}}{\text{)}}_2(s)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g ${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}(s)$

Number of moles can be calculated as follows;

.

Interpretation Introduction

(g)

Interpretation:

The number of moles in 2.45 g of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}(g)$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{ amu}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.0556{\text{ mole of C}}_{\text{3}}{\text{H}}_{\text{8}}(g)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}(g)$

Calculation:

Number of moles can be calculated as follows;

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}\\ =\frac{\text{2}\text{.45 g}}{\text{44}\text{.1 g/mol}}\\ =0.0556{\text{ mole of C}}_{\text{3}}{\text{H}}_{\text{8}}(g)\end{array}$.

Interpretation Introduction

(h)

Interpretation:

The number of moles in 2.45 g of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}(s)$ should be calculated.

Concept Introduction:

The total number of protons and number of neutrons of any element is called, atomic mass of that element. It is a decimal digit for example the atomic mass of $\text{H}-\text{1 }$ is $1.008\text{a}\text{m}\text{u}$.

The molar mass of any element is the mass of $6.02\times {10}^{23}$ atoms or 1 mole of that element. The unit of molar mass is grams.

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 15CR

There are $0.00716{\text{ mole of Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3(s)$ in 2.45 g.

Explanation of Solution

Given Information:

2.45 g ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3(s)$

Number of moles can be calculated as follows;

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}\\ =\frac{\text{2}\text{.45 g}}{\text{342}\text{.15 g/mol}}\\ =0.00716{\text{ mole of Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}(s)\end{array}$.