Chapter 21, Problem 59A

Interpretation Introduction

Interpretation: The given table needs to be completed for the electrolysis of water.

Concept Introduction: At STP, the relation between the number of moles, pressure, volume, and temperature is as follows:

  $PV=nRT$

Here, the number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Explanation of Solution

In the table, the amount of ${\text{H}}_{\text{2}}\text{O}$ used, ${\text{H}}_{\text{2}}$ formed, and ${\text{O}}_{\text{2}}$ formed is given. The overall reaction for the electrolysis of water is as follows:

  $2{\text{ H}}_{\text{2}}\text{O}(l)\to 2{\text{ H}}_{\text{2}}\text{(}g)+{\text{ O}}_{\text{2}}(g)$

From the above reaction, 2 mol of ${\text{H}}_{\text{2}}\text{O}$ gives 2 mol of ${\text{H}}_{\text{2}}$ and 1 mol of ${\text{O}}_{\text{2}}$ . Thus, if the amount of any one species is known, the amount for another can be calculated by considering the correct stoichiometric coefficient.

In (a), the number of moles of water used is given 2.0 mol, and the number of moles of hydrogen gas and oxygen gas needs to be calculated.

From the above reaction, 2 mol of water gives 2 mol of hydrogen gas and 1 mol of oxygen gas thus, the number of moles of hydrogen gas and oxygen gas formed will be 2 mol and 1 mol respectively.

(b)

The mass of oxygen gas formed is 16.0 g. Now, the molar mass of oxygen gas is 32 g/mol; thus, the number of moles of oxygen can be calculated as follows:

  $n=\frac{m}{M}$

Substitute the values,

  $\begin{array}{c}n=\frac{16\text{ g}}{32\text{ g/mol}}\\ =0.5\text{ mol}\end{array}$

Thus, the number of moles of oxygen gas is 0.5 mol.

From the reaction of electrolysis of water, 1 mol of oxygen gas is formed from 2 mol of water thus, the number of moles of water used to form 0.5 mol of oxygen gas will be:

  $n=2\times 0.5=1\text{ mol}$

Now, 2 mol of water formed 2 mol of hydrogen gas thus, 1 mol of water forms 1 mol of hydrogen gas.

The number of hydrogen gas formed and water used is 1 mol and 2 mol respectively.

The molar mass of hydrogen gas is 2 g/mol and that of water is 18 g/mol thus, the mass of hydrogen gas and water will be:

  $\begin{array}{l}n=\frac{m}{M}\\ \text{Or,}\\ m=n\times M\end{array}$

Substitute the values,

  $\begin{array}{c}{m}_{{\text{H}}_2}=\left(1\text{ mol}\right)\left(2\text{ g/mol}\right)\\ =2\text{ g}\\ m_{{\text{H}}_{\text{2}}\text{O}}=\left(2\text{ mol}\right)\left(18\text{ g/mol}\right)\\ =36\text{ g}\end{array}$

Thus, the mass of hydrogen gas and water is 2 g and 36 g respectively.

(c)

The mass of hydrogen gas formed is given 10.0 g. The molar mass of hydrogen gas is 2 g/mol. The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Putting the values,

  $\begin{array}{c}n=\frac{10.0\text{ g}}{2\text{ g/mol}}\\ =5\text{ mol}\end{array}$

Now, for 2 mol of hydrogen gas formed, 1 mol of oxygen gas is formed. The number of moles of oxygen gas formed in this case will be 2.5 mol.

Molar mass of oxygen gas is 32 g/mol; thus, the mass of oxygen gas formed will be:

  $m=n\times M$

Substituting the values,

  $m=\left(2.5\text{ mol}\right)\left(32\text{ g/mol}\right)=80\text{ g}$

Thus, the mass of oxygen gas formed will be 80 g.

Now, from the reaction, 2 mol of water is used to give 2 mol of hydrogen gas; thus, for 5 mol of hydrogen gas formed, the number of moles of water used will be 5 mol.

From the number of moles, the mass of water can be calculated as follows:

  $m=n\times M$

Or,

  $\begin{array}{c}m=\left(5\text{ mol}\right)\left(18\text{ g/mol}\right)\\ =90\text{ g}\end{array}$

Since the density of water is approx. 1 g/mL, the volume of water in mL will be:

  $\begin{array}{c}d=\frac{m}{V}\\ V=\frac{m}{d}\\ =\frac{90\text{ g}}{1\text{ g/mL}}\\ =90\text{ mL}\end{array}$

Thus, the volume of water is 90 mL.

(d)

The mass of water is given 44.4 g. The molar mass of water is 18 g/mol thus, the number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Substituting the values,

  $\begin{array}{c}n=\frac{44.4\text{ g}}{18\text{ g/mol}}\\ =2.46\text{ mol}\end{array}$

From the reaction 2 mol of water gives 2 mol of hydrogen gas and 1 mol of oxygen gas thus, the number of moles of hydrogen gas formed will be 2.46 mol and that of oxygen gas will be as follows:

  $\begin{array}{c}n=\frac{2.46}{2}\text{ mol}\\ =1.23\text{ mol}\end{array}$

Now, the molar mass of hydrogen gas is 2 g/mol and that of oxygen gas is 32 g/mol. The mass of hydrogen gas and oxygen gas can be calculated as follows:

  $m=n\times M$

Substitute the values,

  $\begin{array}{c}{m}_{{\text{H}}_{\text{2}}}=\left(2.46\text{ mol}\right)\left(2\text{ g/mol}\right)\\ =4.92\text{ g}\end{array}$

Similarly,

  $\begin{array}{c}{m}_{{\text{O}}_{\text{2}}}=\left(1.23\text{ mol}\right)\left(32\text{ g/mol}\right)\\ =39.36\text{ g}\end{array}$

The mass of hydrogen gas and oxygen gas is 4.92 g and 39.36 g respectively.

(e)

The volume of hydrogen gas formed is 8.80 L at STP, the number of moles can be calculated as follows:

  $PV=nRT$

Or,

  $n=\frac{PV}{RT}$

Substituting the values,

  $\begin{array}{c}n=\frac{\left(1\text{ atm}\right)\left(8.80\text{ L}\right)}{\left(0.0832\text{ L atm/K mol}\right)\left(273\text{ K}\right)}\\ =0.387\text{ mol}\end{array}$

Now, from the electrolysis reaction of water, 2 mol of hydrogen gas is formed from 2 mol of water thus, the number of moles of water used to form 0.387 mol of hydrogen gas will be 0.387 mol.

The molar mass of water is 18 g/mol; thus, the mass of water used will be:

  $\begin{array}{c}m=n\times M\\ =\left(0.387\text{ mol}\right)\left(18\text{ g/mol}\right)\\ =6.966\text{ g}\end{array}$

Similarly, for 2 mol of hydrogen gas formed, the number of moles of oxygen gas formed is 1 mol.

The number of moles of oxygen gas formed for 0.387 mol of hydrogen gas will be:

  $\begin{array}{c}n=\frac{0.387\text{ mol}}{2}\\ =0.194\text{ mol}\end{array}$

At STP, the volume of oxygen gas formed will be:

  $PV=nRT$

Or,

  $V=\frac{nRT}{P}$

Substituting the values,

  $\begin{array}{c}V=\frac{\left(\text{0}\text{.194 mol}\right)\left(0.0832\text{ L atm/K mol}\right)\left(273\text{ K}\right)}{\left(1\text{ atm}\right)}\\ =4.40\text{ L}\end{array}$

Thus, the volume of oxygen gas formed will be $4.40\text{ L}$ .

(f)

The volume of water used is 66.0 mL. Now, the density of water is 1 g/mL thus, the mass of water will be 66 g.

The number of moles of water can be calculated as follows:

  $n=\frac{m}{M}$

Or,

  $\begin{array}{c}n=\frac{66\text{ g}}{\text{18 g/mol}}\\ =3.67\text{ mol}\end{array}$

From the reaction, 2 mol of water gives 2 mol of hydrogen gas and 1 mol of oxygen gas.

The number of moles of hydrogen gas formed will be 3.67 mol and that of oxygen gas will be:

  $\begin{array}{c}n=\frac{3.67}{2}\\ =1.835\text{ mol}\end{array}$

The mass of hydrogen gas formed can be calculated as follows:

  $\begin{array}{c}m=n\times M\\ =\left(3.67\text{ mol}\right)\left(2\text{ g/mol}\right)\\ =7.34\text{ g}\end{array}$

At STP, the volume of oxygen gas formed will be:

  $\begin{array}{l}PV=nRT\\ \text{Or,}\\ V=\frac{nRT}{P}\end{array}$

Substitute the values,

  $\begin{array}{c}V=\frac{\left(\text{1}\text{.835 mol}\right)\left(0.0832\text{ L atm/K mol}\right)\left(273\text{ K}\right)}{\left(1\text{ atm}\right)}\\ =41.68\text{ L}\end{array}$

Thus, the volume of oxygen gas formed is $41.68\text{ L}$ .

The complete data table will be as follows:

${\text{H}}_{\text{2}}\text{O}$ used	${\text{H}}_{\text{2}}\text{ formed}$	${\text{O}}_{\text{2}}\text{ formed}$
2.0 mol	2.0 mol	1.0 mol
36 g	2 g	16.0 g
90 mL	10.0 g	80 g
44.4 g	4.92 g	39.36 g
6.966 g	8.80 L (STP)	4.40 L
66.0 mL	7.34 g	41.68 L