Chapter 5, Problem 5.79P

How many grams are contained in 5.00 mol of each compound?

1. HCl
2. Na₂ SO₄
3. C₂H₂
4. Al(OH)₃

Interpretation Introduction

(a)

Interpretation:

The amount of $5.00$ mol of HCl is to be determined in grams .

Concept Introduction:

Following is the formula that will be used to calculate the mass of the substance:

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

Answer to Problem 5.79P

The amount of $\text{HCl}$ present in $5.00$ mol of $\text{HCl}$ is $182.3g$.

Explanation of Solution

Number of moles of $\text{HCl}$

  $\text{= 5}\text{.00 mol}$

Molar mass of $\text{HCl}$

  $\text{= 36}\text{.46 g/mol}$

The formula will be used to calculate the mass of $\text{HCl}$ −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

  $mass=5.00mol\times 36.46g/mol$

  $mass=182.3g$

The amount of $\text{HCl}$ present in $5.00$ mol of $\text{HCl}$ is $182.3g$.

Interpretation Introduction

(b)

Interpretation:

The amount of $5.00$ mol of ${\text{Na}}_2{\text{SO}}_4$ is to be determined in grams .

Concept Introduction:

Following is the formula that will be used to calculate the mass of the substance:

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

Answer to Problem 5.79P

The amount of ${\text{Na}}_2{\text{SO}}_4$ present in $5.00$ mol of ${\text{Na}}_2{\text{SO}}_4$ is $710g$

Explanation of Solution

Number of moles of ${\text{Na}}_2{\text{SO}}_4$

  $\text{= 5}\text{.00 mol}$

Molar mass of ${\text{Na}}_2{\text{SO}}_4$

  $\text{= 142}\text{.05 g/mol}$

The formula will be used to calculate the mass of ${\text{Na}}_2{\text{SO}}_4$ −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

  $mass=5.00mol\times 142.05g/mol$

$mass=182.3g$

The amount of ${\text{Na}}_2{\text{SO}}_4$ present in $5.00$ mol of ${\text{Na}}_2{\text{SO}}_4$ is $710g$

Interpretation Introduction

(c)

Interpretation:

The amount of $5.00$ mol of ${\text{C}}_2{\text{H}}_2$ is to be determined in grams .

Concept Introduction:

Following is the formula that will be used to calculate the mass of the substance:

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

Answer to Problem 5.79P

The amount of ${\text{C}}_2{\text{H}}_2$ present in $5.00$ mol of ${\text{C}}_2{\text{H}}_2$ is $130g$.

Explanation of Solution

Number of moles of ${\text{C}}_2{\text{H}}_2$

  $\text{= 5}\text{.00 mol}$

Molar mass of ${\text{C}}_2{\text{H}}_2$

  $\text{= 26}\text{.04 g/mol}$

The formula will be used to calculate the mass of ${\text{C}}_2{\text{H}}_2$ −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

  $mass=5.00mol\times 26.04g/mol$

  $mass=130g$

The amount of ${\text{C}}_2{\text{H}}_2$ present in $5.00$ mol of ${\text{C}}_2{\text{H}}_2$ is $130g$.

Interpretation Introduction

(d)

Interpretation:

The amount of $5.00$ mol of ${\text{Al(OH)}}_3$ is to be determined in grams .

Concept Introduction:

Following is the formula that will be used to calculate the mass of the substance:

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

Answer to Problem 5.79P

The amount of ${\text{Al(OH)}}_3$ present in $5.00$ mol of ${\text{Al(OH)}}_3$ is $390g$.

Explanation of Solution

Number of moles of ${\text{Al(OH)}}_3$

  $\text{= 5}\text{.00 mol}$

Molar mass of ${\text{Al(OH)}}_3$

  $\text{= 78}\text{.03 g/mol}$

The formula will be used to calculate the mass of ${\text{Al(OH)}}_3$ −

  $numberofmoles=\frac{mass}{Molarmass}$

Or,

  $mass=numberofmoles\times Molarmass$

  $mass=5.00mol\times 78.03g/mol$

The amount of ${\text{Al(OH)}}_3$ present in $5.00$ mol of ${\text{Al(OH)}}_3$ is $390g$.