Chapter 11.1, Problem 3PP

(a)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. The possible mole ratio has to be determined.

$\text{4Al}\left(s\right)+3{\text{O}}_2\left(aq\right)\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Concept introduction:

Mole ratio is use for the conversion between products and reactants. It is defined as the ratio of moles of any of the two compounds that are involved in the reaction.

Answer to Problem 3PP

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{4\text{mol Al}}{3\text{mol}{\text{O}}_2}\text{ and }\frac{4\text{mol Al}}{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}\\ \frac{3{\text{mol O}}_2}{4\text{mol}\text{Al}}\text{ and }\frac{3{\text{mol O}}_2}{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}\\ \frac{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}{4\text{mol}\text{Al}}\text{ and }\frac{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}{3\text{mol}{\text{O}}_2}\end{array}$

Explanation of Solution

The balanced chemical equation is shown below.

$\text{4Al}\left(s\right)+3{\text{O}}_2\left(aq\right)\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}$

The above equation shows that four moles of aluminum reacts with three moles of oxygen to form 2 moles of aluminum oxide.

Therefore, the mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{4\text{mol Al}}{3\text{mol}{\text{O}}_2}\text{ and }\frac{4\text{mol Al}}{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}\\ \frac{3{\text{mol O}}_2}{4\text{mol}\text{Al}}\text{ and }\frac{3{\text{mol O}}_2}{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}\\ \frac{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}{4\text{mol}\text{Al}}\text{ and }\frac{2\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_3}{3\text{mol}{\text{O}}_2}\end{array}$

(b)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. The possible mole ratio has to be determined.

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

Concept introduction:

Mole ratio is use for the conversion between products and reactants. It is defined as the ratio of moles of any of the two compounds that are involved in the reaction.

Answer to Problem 3PP

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{3\text{mol}\text{Fe}}{3\text{mol}{\text{H}}_{\text{2}}\text{O}},\frac{3\text{mol}\text{Fe}}{1\text{mol}{\text{Fe}}_3{\text{O}}_4}\text{and}\frac{3\text{mol}\text{Fe}}{4\text{mol}{\text{H}}_{\text{2}}}\\ \frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{3\text{mol}\text{Fe}},\frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{Fe}}_3{\text{O}}_4}\text{and}\frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{4\text{mol}{\text{H}}_{\text{2}}}\\ \frac{1\text{mol}{\text{Fe}}_3{\text{O}}_4}{3\text{mol}\text{Fe}},\frac{1\text{mol}{\text{Fe}}_3{\text{O}}_4}{4\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{1\text{mol}{\text{Fe}}_3{\text{O}}_4}{4\text{mol}{\text{H}}_{\text{2}}}\\ \frac{4\text{mol}{\text{H}}_{\text{2}}}{3\text{mol}\text{Fe}},\frac{4\text{mol}{\text{H}}_{\text{2}}}{4\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{4\text{mol}{\text{H}}_{\text{2}}}{1\text{mol}{\text{Fe}}_3{\text{O}}_4}\end{array}$

Explanation of Solution

The balanced chemical equation is shown below.

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

The above equation shows that three moles of iron reacts with four moles of water to form 1 mole of ferric oxide and four moles of hydrogen.

Therefore, the mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{3\text{mol}\text{Fe}}{3\text{mol}{\text{H}}_{\text{2}}\text{O}},\frac{3\text{mol}\text{Fe}}{1\text{mol}{\text{Fe}}_3{\text{O}}_4}\text{and}\frac{3\text{mol}\text{Fe}}{4\text{mol}{\text{H}}_{\text{2}}}\\ \frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{3\text{mol}\text{Fe}},\frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{1\text{mol}{\text{Fe}}_3{\text{O}}_4}\text{and}\frac{4\text{mol}{\text{H}}_{\text{2}}\text{O}}{4\text{mol}{\text{H}}_{\text{2}}}\\ \frac{1\text{mol}{\text{Fe}}_3{\text{O}}_4}{3\text{mol}\text{Fe}},\frac{1\text{mol}{\text{Fe}}_3{\text{O}}_4}{4\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{1\text{mol}{\text{Fe}}_3{\text{O}}_4}{4\text{mol}{\text{H}}_{\text{2}}}\\ \frac{4\text{mol}{\text{H}}_{\text{2}}}{3\text{mol}\text{Fe}},\frac{4\text{mol}{\text{H}}_{\text{2}}}{4\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{4\text{mol}{\text{H}}_{\text{2}}}{1\text{mol}{\text{Fe}}_3{\text{O}}_4}\end{array}$

(c)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. The possible mole ratio has to be determined.

$2\text{HgO}\left(s\right)\to \text{2Hg}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

Concept introduction:

Mole ratio is use for the conversion between products and reactants. It is defined as the ratio of moles of any of the two compounds that are involved in the reaction.

Answer to Problem 3PP

The mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{2\text{mol HgO}}{2\text{mol Hg}}\text{ and }\frac{2\text{mol HgO}}{1\text{mol}{\text{O}}_2}\\ \frac{2\text{mol Hg}}{2\text{mol HgO}}\text{ and }\frac{3\text{mol Hg}}{1\text{mol}{\text{O}}_2}\\ \frac{1\text{mol}{\text{O}}_2}{2\text{mol HgO}}\text{ and }\frac{1\text{mol}{\text{O}}_2}{2\text{mol Hg}}\end{array}$

Explanation of Solution

The balanced chemical equation is shown below.

$2\text{HgO}\left(s\right)\to \text{2Hg}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)$

The above equation shows that 2 moles of mercuric oxide on decomposition gives two moles of mercury and one mole of oxygen.

Therefore, the mole ratio for given chemical equation is shown below.

$\begin{array}{l}\frac{2\text{mol HgO}}{2\text{mol Hg}}\text{ and }\frac{2\text{mol HgO}}{1\text{mol}{\text{O}}_2}\\ \frac{2\text{mol Hg}}{2\text{mol HgO}}\text{ and }\frac{3\text{mol Hg}}{1\text{mol}{\text{O}}_2}\\ \frac{1\text{mol}{\text{O}}_2}{2\text{mol HgO}}\text{ and }\frac{1\text{mol}{\text{O}}_2}{2\text{mol Hg}}\end{array}$