Chapter 14, Problem 98A

Interpretation Introduction

Interpretation: Use the given data to calculate the mole ratio for the reaction between ${\text{KNO}}_{\text{3}}$ and ${\text{O}}_{\text{2}}$

Concept Introduction: Pressure is related to the number of moles, temperature and volume as follows:

  $PV=nRT$

Here, P is pressure, V is volume, n is the number of moles, R is the Universal gas constant and T is temperature.

Explanation of Solution

When potassium nitrate is heated, oxygen gas is produced. The volumes of oxygen produced at STP from different masses of potassium nitrate are given in the table below:

Mass of ${\text{KNO}}_{\text{3}}$ (g)	Volume of ${\text{O}}_{\text{2}}$ (cL)
0.84	9.3
1.36	15.1
2.77	30.7
4.82	53.5
6.96	77.3

From the given masses of potassium nitrate, the number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Molar mass of potassium nitrate is 101.10 g/mol.

Also, the given volume of oxygen is produced at STP. Thus, the temperature will be 273.15 K and the pressure will be 1 atm.

Now, the number of moles can be calculated in each case as follows:

  $n=\frac{P\text{}V}{RT}$

Mass of ${\text{KNO}}_{\text{3}}$ (g)	Number of moles ${\text{KNO}}_{\text{3}}$ (mol)	Volume of ${\text{O}}_{\text{2}}$ (cL)	Number of moles ${\text{O}}_{\text{2}}$ (mol)
0.84	$\begin{array}{c}{n}_{{\text{KNO}}_{\text{3}}}=\frac{0.84}{101.10}\\ =0.0083\text{ mol}\end{array}$	9.3	$\begin{array}{c}n=\frac{\left(1\text{ atm}\right)\left(9.3\text{ cL}\right)\left(\frac{0.01\text{ L}}{1\text{ cL}}\right)}{\left(0.082\text{ L atm/mol K}\right)\left(273.15\text{ K}\right)}\\ =\frac{0.093}{22.4}\\ =0.0042\text{ mol}\end{array}$
1.36	$\begin{array}{c}{n}_{{\text{KNO}}_{\text{3}}}=\frac{1.36}{101.10}\\ =0.0135\text{ mol}\end{array}$	15.1	$\begin{array}{c}n=\frac{\left(1\text{ atm}\right)\left(15.1\text{ cL}\right)\left(\frac{0.01\text{ L}}{1\text{ cL}}\right)}{\left(0.082\text{ L atm/mol K}\right)\left(273.15\text{ K}\right)}\\ =0.00674\text{ mol}\end{array}$
2.77	$\begin{array}{c}{n}_{{\text{KNO}}_{\text{3}}}=\frac{2.77}{101.10}\\ =0.03\text{ mol}\end{array}$	30.7	$\begin{array}{c}n=\frac{\left(1\text{ atm}\right)\left(30.7\text{ cL}\right)\left(\frac{0.01\text{ L}}{1\text{ cL}}\right)}{\left(0.082\text{ L atm/mol K}\right)\left(273.15\text{ K}\right)}\\ =\frac{0.093}{22.4}\\ =0.014\text{ mol}\end{array}$
4.82	$\begin{array}{c}{n}_{{\text{KNO}}_{\text{3}}}=\frac{4.82}{101.10}\\ =0.048\text{ mol}\end{array}$	53.5	$\begin{array}{c}n=\frac{\left(1\text{ atm}\right)\left(53.5\text{ cL}\right)\left(\frac{0.01\text{ L}}{1\text{ cL}}\right)}{\left(0.082\text{ L atm/mol K}\right)\left(273.15\text{ K}\right)}\\ =0.024\text{ mol}\end{array}$
6.96	$\begin{array}{c}{n}_{{\text{KNO}}_{\text{3}}}=\frac{6.96}{101.10}\\ =0.068\text{ mol}\end{array}$	77.3	$\begin{array}{c}n=\frac{\left(1\text{ atm}\right)\left(77.3\text{ cL}\right)\left(\frac{0.01\text{ L}}{1\text{ cL}}\right)}{\left(0.082\text{ L atm/mol K}\right)\left(273.15\text{ K}\right)}\\ =0.035\text{ mol}\end{array}$

From the above table, the mole ratio of ${\text{KNO}}_{\text{3}}$ and ${\text{O}}_{\text{2}}$ can be estimated as follows:

Number of moles ${\text{KNO}}_{\text{3}}$ (mol)	Number of moles ${\text{O}}_{\text{2}}$ (mol)	Mole ratio $\left({\text{KNO}}_{\text{3}}{\text{/O}}_{\text{2}}\right)$
$0.0083\text{ mol}$	$0.0042\text{ mol}$	$\frac{0.0083}{0.0042}=1.97\approx 2$
$0.0135\text{ mol}$	$0.00674\text{ mol}$	$\frac{0.0135}{0.00674}=2.003\approx 2$
$0.03\text{ mol}$	$0.014\text{ mol}$	$\frac{0.03}{0.014}=2.14\approx 2$
$0.048\text{ mol}$	$0.024\text{ mol}$	$\frac{0.048}{0.024}=2$
$0.068\text{ mol}$	$0.035\text{ mol}$	$\frac{0.068}{0.035}=1.94\approx 2$

From the above calculations, the mole ratio for the reaction between ${\text{KNO}}_{\text{3}}$ and ${\text{O}}_{\text{2}}$ is 2:1.