Chapter 6, Problem 26QP

(a)

Interpretation Introduction

Interpretation:

The given equation ${\text{C}}_6{\text{H}}_{12}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to C{O}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$ is to be balanced and after that, the number of grams of the first reactant that is required to react completely to give $1.70\text{ g}$ of the product indicated in the boldface is to be determined.

Explanation of Solution

The relation among the number of moles, molar mass and mass is,

$n=\frac{m}{\text{MM}}$

Here, $n$ is the number of moles, $m$ is mass and $\text{MM}$ is molar mass.

By rearranging the above equation mass can be determined as follows:

$m=n\times \text{MM}$ …..(1)

The balanced equation is,

${\text{C}}_6{\text{H}}_{12}\left(l\right)+9{\text{O}}_{\text{2}}\left(g\right)\to 6C{O}_2\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$

The molar mass of ${\text{CO}}_2$ is $44.01\text{g/mol}$ and ${\text{C}}_6{\text{H}}_{12}$ is $84.16\text{g/mol}$ . The mass of ${\text{C}}_6{\text{H}}_{12}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=1.70\cancel{{\text{g CO}}_2}\times \frac{\cancel{{\text{mol CO}}_2}}{44.01\cancel{{\text{g CO}}_2}}\times \frac{1\cancel{{\text{mol C}}_6{\text{H}}_{12}}}{6\cancel{{\text{mol CO}}_2}}\times \frac{84.16{\text{g C}}_6{\text{H}}_{12}}{1\cancel{{\text{mol C}}_6{\text{H}}_{12}}}\\ =0.542{\text{ g C}}_6{\text{H}}_{12}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The given equation ${\text{NI}}_3\left(s\right)\to {\text{N}}_{\text{2}}\left(g\right)+I_2\left(s\right)$ is to be balanced and after that, the number of grams of the first reactant that is required to react completely to give $1.70\text{ g}$ of the product indicated in the boldface is to be determined.

Explanation of Solution

The balanced equation is,

${\text{2NI}}_3\left(s\right)\to {\text{N}}_{\text{2}}\left(g\right)+3I_2\left(s\right)$

The molar mass of ${\text{I}}_2$ is $253.8\text{g/mol}$ and ${\text{NI}}_3$ is $394.7\text{g/mol}$ . The mass of ${\text{NI}}_3$ is determined by using equation (1) as follows:

$\begin{array}{c}m=1.70\cancel{{\text{g I}}_2}\times \frac{\cancel{{\text{mol I}}_2}}{253.8\cancel{{\text{g I}}_2}}\times \frac{2\cancel{{\text{mol NI}}_3}}{3\cancel{{\text{mol I}}_2}}\times \frac{394.7{\text{g NI}}_3}{1\cancel{{\text{mol NI}}_3}}\\ =1.76{\text{ g NI}}_3\end{array}$

(c)

Interpretation Introduction

Interpretation:

The given equation $\text{Na}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to N_{2}O\left(s\right)$ is to be balanced and after that, the number of grams of the first reactant that is required to react completely to give $1.70\text{ g}$ of the product indicated in the boldface is to be determined.

Explanation of Solution

The balanced equation is,

$\text{4Na}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2N{a}_{2}O\left(s\right)$

The molar mass of ${\text{Na}}_2\text{O}$ is $61.98\text{g/mol}$ and $\text{Na}$ is $22.99\text{g/mol}$ . The mass of $\text{Na}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=1.70\cancel{{\text{g Na}}_2\text{O}}\times \frac{\cancel{{\text{mol Na}}_2\text{O}}}{61.98\cancel{{\text{g Na}}_2\text{O}}}\times \frac{4\cancel{\text{mol Na}}}{2\cancel{{\text{mol Na}}_2\text{O}}}\times \frac{22.99\text{g Na}}{1\cancel{\text{mol Na}}}\\ =1.26\text{ g Na}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The given equation $\text{Ca}\left(s\right)+\text{HCl}\left(aq\right)\to {\text{CaCl}}_2\left(aq\right)+H_2\left(g\right)$ is to be balanced and after that, the number of grams of the first reactant that is required to react completely to give $1.70\text{ g}$ of the product indicated in the boldface is to be determined.

Explanation of Solution

The balanced equation is,

$\text{Ca}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{CaCl}}_2\left(aq\right)+H_2\left(g\right)$

The molar mass of ${\text{H}}_2$ is $2.016\text{g/mol}$ and $\text{Ca}$ is $40.08\text{g/mol}$ . The mass of $\text{Ca}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=1.70\cancel{{\text{g H}}_2}\times \frac{\cancel{{\text{mol H}}_2}}{2.016\cancel{{\text{g H}}_2}}\times \frac{1\cancel{\text{mol Ca}}}{1\cancel{{\text{mol H}}_2}}\times \frac{40.08\text{g Ca}}{1\cancel{\text{mol Ca}}}\\ =33.8\text{ g Ca}\end{array}$