Chapter 5, Problem 75P

Interpretation Introduction

(a)

Interpretation:

Number of grams of ${\text{CO}}_2$ that is formed from $\text{2}\text{.5 mol}$ of ${\text{C}}_2{\text{H}}_2$ is to be determined considering the following combustion reaction of acetylene:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Concept Introduction:

The carbon-containing compound i.e., acetylene, reacts with oxygen to form water and carbon dioxide. This reaction is known as combustion.

The combustion reaction of acetylene ${\text{C}}_2{\text{H}}_2$ is as shown below:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$.

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

  $mass=numberofmoles\times Molarmass$

Answer to Problem 75P

  $\text{220}{\text{.5g of CO}}_2\text{ are formed from 2}{\text{.5 mol of C}}_{\text{2}}{\text{H}}_{\text{2}}$

Explanation of Solution

According to the equation: ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Moles ratio of ${\text{C}}_2{\text{H}}_2$ $:{\text{CO}}_2$ $=\text{ 2:4}$ ,

i.e., if $\text{2}$ moles of ${\text{C}}_2{\text{H}}_2$ requires $4$ mole of ${\text{CO}}_2$ ,

  $\begin{array}{l}\text{then 2}{\text{.5mole of C}}_{\text{2}}{\text{H}}_2\text{will requir}e=\frac{2.5\times 4}{2}{\text{ moles of CO}}_2\\ ={\text{ 5 moles of CO}}_2\end{array}$

  $mass=numberofmoles\times Molarmass$

  ${\text{molar mass of CO}}_2=44.01g/mol$

  $\begin{array}{l}mass=\text{ 5}\text{.0 mol}\times 44.01g/mol\\ \text{ =220}\text{.05 g}\end{array}$

Hence, 220.5 g of ${\text{CO}}_2$ is formed from 2.5 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$.

Interpretation Introduction

(b)

Interpretation:

Number of grams of ${\text{CO}}_2$ that is formed from $\text{0}\text{.5 mol}$ of ${\text{C}}_2{\text{H}}_2$ is to be determined considering the following combustion reaction of acetylene:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Concept Introduction:

The carbon-containing compound i.e., acetylene, reacts with oxygen to form water and carbon dioxide. This reaction is known as combustion.

The combustion reaction of acetylene ${\text{C}}_2{\text{H}}_2$ is as shown below:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$.

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

  $mass=numberofmoles\times Molarmass$

Answer to Problem 75P

  $44.01{\text{ g of CO}}_2\text{ are formed from 0}{\text{.5 mol of C}}_{\text{2}}{\text{H}}_{\text{2}}$

Explanation of Solution

According to the equation: ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Moles ratio of ${\text{C}}_2{\text{H}}_2$ $:{\text{CO}}_2$ $=\text{ 2:4}$ ,

i.e., if $\text{2}$ moles of ${\text{C}}_2{\text{H}}_2$ requires $4$ mole of ${\text{CO}}_2$ ,

  $\begin{array}{l}\text{then 0}{\text{.5 mole of C}}_{\text{2}}{\text{H}}_2\text{will requir}e=\frac{0.5\times 4}{2}{\text{ moles of CO}}_2\\ =\text{ 1}{\text{.0 moles of CO}}_2\end{array}$

  $mass=numberofmoles\times Molarmass$

  ${\text{molar mass of CO}}_2=44.01g/mol$

  $\begin{array}{l}mass=\text{ 1 mol}\times 44.01g/mol\\ \text{ =}44.01\text{ g}\end{array}$

Hence, 44.01 g of ${\text{CO}}_2$ is formed from 0.5 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$.

Interpretation Introduction

(c)

Interpretation:

Number of grams of ${\text{H}}_2\text{O}$ that is formed from $\text{0}\text{.25 mol}$ of ${\text{C}}_2{\text{H}}_2$ is to be determined considering the following combustion reaction of acetylene:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Concept Introduction:

The carbon-containing compound i.e., acetylene, reacts with oxygen to form water and carbon dioxide. This reaction is known as combustion.

The combustion reaction of acetylene ${\text{C}}_2{\text{H}}_2$ is as shown below:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$.

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

  $mass=numberofmoles\times Molarmass$

Answer to Problem 75P

  $\text{4}{\text{.505 g of H}}_2\text{O are formed from 0}{\text{.25 mol of C}}_{\text{2}}{\text{H}}_{\text{2}}$

Explanation of Solution

According to the equation: ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Moles ratio of ${\text{C}}_2{\text{H}}_2$ $:{\text{H}}_2\text{O}$ $=\text{ 2:2}$ ,

i.e., if $\text{2}$ moles of ${\text{C}}_2{\text{H}}_2$ requires $2$ mole of ${\text{H}}_2\text{O}$ ,

  $\begin{array}{l}\text{then 0}{\text{.5 mole of C}}_{\text{2}}{\text{H}}_2\text{will requir}e=\frac{0.25\times 2}{2}{\text{ moles of H}}_2\text{O}\\ =\text{ 0}{\text{.25 moles of H}}_2\text{O}\end{array}$

  $mass=numberofmoles\times Molarmass$

  ${\text{molar mass of H}}_2\text{O}=18.02g/mol$

  $\begin{array}{l}mass=\text{ 0}\text{.25 mol}\times 18.02g/mol\\ \text{ =}4.505\text{ g}\end{array}$

Hence, 4.505 g of ${\text{H}}_2\text{O}$ is formed from 0.25 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$.

Interpretation Introduction

(d)

Interpretation:

Number of grams of ${\text{O}}_2$ that reacts with $\text{3}\text{.0 mol}$ of ${\text{C}}_2{\text{H}}_2$ is to be determined considering the following combustion reaction of acetylene:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Concept Introduction:

The carbon-containing compound i.e., acetylene, reacts with oxygen to form water and carbon dioxide. This reaction is known as combustion.

The combustion reaction of acetylene ${\text{C}}_2{\text{H}}_2$ is as shown below:

  ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$.

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

  $mass=numberofmoles\times Molarmass$

Answer to Problem 75P

  ${\text{120 g of O}}_2{\text{ reacts with 3 mol of C}}_{\text{2}}{\text{H}}_{\text{2}}$.

Explanation of Solution

According to the equation: ${\text{2C}}_2{\text{H}}_2{\text{+5O}}_2\to 4{\text{CO}}_2+2{\text{H}}_2\text{O}$

Moles ratio of ${\text{C}}_2{\text{H}}_2$ $:{\text{O}}_2$ $=\text{ 2:5}$ ,

i.e., if $\text{2}$ moles of ${\text{C}}_2{\text{H}}_2$ requires $5$ mole of ${\text{O}}_2$ ,

  $\begin{array}{l}{\text{then 3 mole of C}}_{\text{2}}{\text{H}}_2\text{will requir}e=\frac{3\times 5}{2}{\text{ moles of O}}_2\\ =\text{ 7}{\text{.5 moles of O}}_2\end{array}$

  $mass=numberofmoles\times Molarmass$

  ${\text{molar mass of O}}_2=16g/mol$

  $\begin{array}{l}mass=\text{ 7}\text{.5 mol}\times 16g/mol\\ \text{ =120 g}\end{array}$

Hence, 120 g of ${\text{O}}_2$ is formed from 3 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$.