Chapter 6, Problem 25QP

(a)

Interpretation Introduction

Interpretation:

The given equation $\text{Cr}\left(s\right)+{\text{Cl}}_{\text{2}}\left(g\right)\to {\text{CrCl}}_3\left(s\right)$ is to be balanced and after that, the number of grams of the second reactant that is required to react completely with $0.600\text{ g}$ of the first reactant is to be determined.

Explanation of Solution

The mass of the first reactant is $0.600\text{ g}$ .

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,

$m=n\times \text{MM}$ …..(1)

The balanced equation is,

$\text{2Cr}\left(s\right)+3{\text{Cl}}_{\text{2}}\left(g\right)\to 2{\text{CrCl}}_3\left(s\right)$

The molar mass of $\text{Cr}$ is $52.00\text{g/mol}$ and ${\text{Cl}}_{\text{2}}$ is $70.90\text{g/mol}$ . The mass of ${\text{Cl}}_{\text{2}}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=0.600\cancel{\text{g Cr}}\times \frac{\cancel{\text{mol Cr}}}{52.00\cancel{\text{g Cr}}}\times \frac{3\cancel{{\text{mol Cl}}_2}}{2\cancel{\text{mol Cr}}}\times \frac{70.90{\text{g Cl}}_2}{1\cancel{{\text{mol Cl}}_2}}\\ =1.23{\text{ g Cl}}_2\end{array}$

(b)

Interpretation Introduction

Interpretation:

The given equation ${\text{RbO}}_2\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_2\left(g\right)+\text{RbOH}\left(s\right)$ is to be balanced and after that, the number of grams of the second reactant that is required to react completely with $0.600\text{ g}$ of the first reactant is to be determined.

Explanation of Solution

The balanced equation is,

${\text{4RbO}}_2\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 3{\text{O}}_2\left(g\right)+4\text{RbOH}\left(s\right)$

The molar mass of ${\text{RbO}}_2$ is $117.47\text{g/mol}$ and ${\text{H}}_{\text{2}}\text{O}$ is $18.02\text{g/mol}$ . The mass of ${\text{H}}_{\text{2}}\text{O}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=0.600\cancel{{\text{g RbO}}_2}\times \frac{\cancel{{\text{mol RbO}}_2}}{117.47\cancel{{\text{g RbO}}_2}}\times \frac{2\cancel{{\text{mol H}}_{\text{2}}\text{O}}}{4\cancel{{\text{mol RbO}}_2}}\times \frac{18.02{\text{g H}}_{\text{2}}\text{O}}{1\cancel{{\text{mol H}}_{\text{2}}\text{O}}}\\ =0.0460{\text{ g H}}_{\text{2}}\text{O}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The given equation ${\text{C}}_5{\text{H}}_{12}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$ is to be balanced and after that, the number of grams of the second reactant that is required to react completely with $0.600\text{ g}$ of the first reactant is to be determined.

Explanation of Solution

The balanced equation is,

${\text{C}}_5{\text{H}}_{12}\left(g\right)+8{\text{O}}_{\text{2}}\left(g\right)\to 5{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$

The molar mass of ${\text{C}}_5{\text{H}}_{12}$ is $72.15\text{g/mol}$ and ${\text{O}}_{\text{2}}$ is $32.00\text{g/mol}$ . The mass of ${\text{O}}_{\text{2}}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=0.600\cancel{{\text{g C}}_5{\text{H}}_{12}}\times \frac{\cancel{{\text{mol C}}_5{\text{H}}_{12}}}{72.15\cancel{{\text{g C}}_5{\text{H}}_{12}}}\times \frac{8\cancel{{\text{mol O}}_{\text{2}}}}{1\cancel{{\text{mol C}}_5{\text{H}}_{12}}}\times \frac{32.00{\text{g O}}_{\text{2}}}{1\cancel{{\text{mol O}}_{\text{2}}}}\\ =2.13{\text{ g O}}_{\text{2}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The given equation $\text{Li}\left(s\right)+{\text{Cl}}_{\text{2}}\left(g\right)\to \text{LiCl}\left(s\right)$ is to be balanced and after that, the number of grams of the second reactant that is required to react completely with $0.600\text{ g}$ of the first reactant is to be determined.

Explanation of Solution

The balanced equation is,

$\text{2Li}\left(s\right)+{\text{Cl}}_{\text{2}}\left(g\right)\to 2\text{LiCl}\left(s\right)$

The molar mass of $\text{Li}$ is $6.94\text{g/mol}$ and ${\text{Cl}}_{\text{2}}$ is $70.90\text{g/mol}$ . The mass of ${\text{O}}_{\text{2}}$ is determined by using equation (1) as follows:

$\begin{array}{c}m=0.600\cancel{\text{g Li}}\times \frac{\cancel{\text{mol Li}}}{6.94\cancel{\text{g Li}}}\times \frac{1\cancel{{\text{mol Cl}}_{\text{2}}}}{2\cancel{\text{mol Li}}}\times \frac{70.90{\text{g Cl}}_{\text{2}}}{1\cancel{{\text{mol Cl}}_{\text{2}}}}\\ =3.06{\text{ g Cl}}_{\text{2}}\end{array}$