Chapter 7, Problem 7.153CQ

(a)

Interpretation Introduction

Interpretation: The number of moles of oxygen gas should be calculated.

Concept Interpretation: The number of moles can be calculated using the following relation:

  $n=\frac{m}{M}$

Here, m is mass in grams and M is molar mass is g/mol.

0.9 mol

The balanced chemical reaction is as follows:

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

From the balanced chemical reaction, 1 mol of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$reacts with 4 mol of oxygen gas thus, 0.225 mole of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$reacts with $4\times 0.225\text{ mol}=\text{0}\text{.9 mol}$of oxygen gas.

(b)

Interpretation Introduction

Interpretation: The mass of water produced should be calculated.

Concept Interpretation: The number of moles can be calculated using the following relation:

  $n=\frac{m}{M}$

Here, m is mass in grams and M is molar mass is g/mol.

The mass can be calculated by rearranging the above reaction as follows:

  $m=n\times M$

18 g

The balanced chemical reaction is as follows:

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

The mass of oxygen gas is 64 g. Molar mass of oxygen gas is 32 g/mol thus, number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Putting the values,

  $n=\frac{64\text{ g}}{32\text{ g/mol}}=2\text{ mol}$

4 mol of oxygen gas produces 2 mol of water thus, 1 mol of oxygen gas produce $\frac{2}{4}\text{ mol}$of water. The number of moles of water produced by 2 mol of oxygen gas will be:

  $n_{H_{2}O}=\frac{2\times 2}{4}\text{ mol}=\text{1mol}$

Molar mass of water is 18 g/mol thus, mass of water can be calculated as follows:

  $m=n\times M$

Putting the values,

  $m=1\text{ mol}\times 18\text{ g/mol=18 g}$

Thus, mass of water is 18 g.

(c)

Interpretation Introduction

Interpretation: The mass of carbon dioxide produced should be calculated.

Concept Interpretation: The number of moles can be calculated using the following relation:

  $n=\frac{m}{M}$

Here, m is mass in grams and M is molar mass is g/mol.

The mass can be calculated by rearranging the above reaction as follows:

  $m=n\times M$

  $\text{257}\text{.05 g}$

The balanced chemical reaction is as follows:

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

The mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$is 78.0 g and molar mass is 40.064 g/mol thus, number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Putting the values,

  $n=\frac{78.0\text{ g}}{40.064\text{ g/mol}}=1.9468\text{ mol}$

1 mol of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$produces 3 mol of carbon dioxide gas thus, 1.9468 mol of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$produces $3\times 1.9468\text{ mol=5}\text{.841 mol}$.

Molar mass of carbon dioxide is 44.01 g/mol thus, mass can be calculated as follows:

  $m=n\times M$

Putting the values,

  $m=5.841\text{ mol}\times 44.01\text{ g/mol=257}\text{.05 g}$

Thus, mass of carbon dioxide is $\text{257}\text{.05 g}$.

(d)

Interpretation Introduction

Interpretation: The percent yield of carbon dioxide should be calculated.

Concept Interpretation: The relation between number of moles and mass of any substance is as follows:

  $n=\frac{m}{M}$

Here, m is mass in grams and M is molar mass is g/mol.

The reactant that limits the amount of product formed in the reaction is known as limiting reactant. The amount of product formed from limiting reactant is known as theoretical yield. The percent yield can be calculated as follows:

  $\%\text{yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\%$

  $72.36\%$

Mass of carbon dioxide is 186 g.

This is the actual yield.

The percentage yield can be calculated as follows:

  $\%\text{yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\%$

Putting the values,

  $\begin{array}{c}\%\text{yield}=\frac{\text{186 g}}{\text{275}\text{.05 g}}\times 100\%\\ =72.36\%\end{array}$

Therefore, percent yield is $72.36\%$.