Chapter 12, Problem 52PE

(a)

Interpretation Introduction

Interpretation:

Volume of oxygen that would react with $7.2{\text{ L C}}_3{\text{H}}_8$ at STP has to be determined.

Concept Introduction:

Avogadro’s law represents direct relation between volume occupied by gas and mole of gas at constant pressure and temperature.

Mathematical relation for Avogadro’s law is as follows:

  $V\propto n$

Here, $n$ is mole of gas and $V$ is volume.

Answer to Problem 52PE

Volume of oxygen that would react with $7.2{\text{ L C}}_3{\text{H}}_8$ is $3.1\text{ L}$.

Explanation of Solution

Chemical equation for reaction between ${\text{C}}_3{\text{H}}_8$ and oxygen is as follows:

  ${\text{C}}_3{\text{H}}_8\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to 3{\text{CO}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(g\right)$

According to Avogadro’s law moles of gas present is directly proportional to volume of gas at constant temperature and pressure.

Volume of ${\text{O}}_2$ required to react with $7.2{\text{ L C}}_3{\text{H}}_8$ can be calculated as follows:

  $\begin{array}{c}\text{Volume}=\left(7.2{\text{ L C}}_3{\text{H}}_8\right)\left(\frac{5{\text{ L O}}_2}{1{\text{ L C}}_3{\text{H}}_8}\right)\\ =36{\text{ L O}}_2\end{array}$

Hence volume of oxygen that would react with $7.2{\text{ L C}}_3{\text{H}}_8$ is $36\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Grams of ${\text{CO}}_2$ that would produce from $35{\text{ L C}}_3{\text{H}}_8$ at STP has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 52PE

Grams of ${\text{CO}}_2$ that would produce from $35{\text{ L C}}_3{\text{H}}_8$ at STP is $31\text{ g}$.

Explanation of Solution

Chemical equation for reaction between ${\text{C}}_3{\text{H}}_8$ and oxygen is as follows:

  ${\text{C}}_3{\text{H}}_8\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to 3{\text{CO}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(g\right)$

According to Avogadro’s law moles of gas present is directly proportional to volume of gas at constant temperature and pressure.

Volume of ${\text{CO}}_2$ that would produce from $35{\text{ L C}}_3{\text{H}}_8$ at can be calculated as follows:

  $\begin{array}{c}\text{Volume}=\left(35{\text{ L C}}_3{\text{H}}_8\right)\left(\frac{3{\text{ L CO}}_2}{1{\text{ L C}}_3{\text{H}}_8}\right)\\ =105{\text{ L CO}}_2\end{array}$

Moles of $105{\text{ L CO}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Mole}\left(\text{mol}\right)=\left(105\text{ L}\right)\left(\frac{1\text{ mol}}{22.4\text{ L}}\right)\\ =4.69\text{ mol}\end{array}$

Grams of $4.69{\text{ mol CO}}_2$ can be calculated as follows:

  $\begin{array}{c}\text{Mass}\left(g\right)=\left(4.69\text{ mol}\right)\left(\frac{44.01\text{ g}}{1\text{ mol}}\right)\\ =210\text{ g}\end{array}$

Hence grams of ${\text{CO}}_2$ that would produce from $35{\text{ L C}}_3{\text{H}}_8$ at STP is $210\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

Volume of ${\text{H}}_2\text{O}$ that would produce from reaction of $15{\text{ L C}}_3{\text{H}}_8$ and $15{\text{ L O}}_2$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 52PE

Volume of ${\text{H}}_2\text{O}$ that would produce from reaction of $15{\text{ L C}}_3{\text{H}}_8$ and $15{\text{ L O}}_2$ is $12\text{ L}$.

Explanation of Solution

Chemical equation for reaction between ${\text{C}}_3{\text{H}}_8$ and oxygen is as follows:

  ${\text{C}}_3{\text{H}}_8\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to 3{\text{CO}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(g\right)$

According to Avogadro’s law moles of gas present is directly proportional to volume of gas at constant temperature and pressure.

Since $1{\text{ L C}}_3{\text{H}}_8$ is required to react with ${\text{5 L O}}_2$ thus $15{\text{ L O}}_2$ in reaction is limiting reagent. Thus amount of product ${\text{H}}_2\text{O}$ is formed according to limiting reagent.

Volume of ${\text{H}}_2\text{O}$ that would produce from $15{\text{ L O}}_2$ at can be calculated as follows:

  $\begin{array}{c}\text{Volume}=\left(15{\text{ L O}}_2\right)\left(\frac{4{\text{ L H}}_2\text{O}}{5{\text{ L O}}_2}\right)\\ =12{\text{ L H}}_2\text{O}\end{array}$

Hence volume of ${\text{H}}_2\text{O}$ that would produce from reaction of $15{\text{ L C}}_3{\text{H}}_8$ and $15{\text{ L O}}_2$ is $12\text{ L}$.