Question

The total pressure of a mixture of oxygen and hydrogen is 1.0 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.40 when measured at the same values of T and V as the original mixture. What is the composition of the original mixture in mole percent?
A.\[{x_{{O_2}}} = 0.2;{x_{{H_2}}} = 0.8\]
B.\[{x_{{O_2}}} = 0.4;{x_{{H_2}}} = 0.6\]
C.\[{x_{{O_2}}} = 0.6;{x_{{H_2}}} = 0.4\]
D.\[{x_{{O_2}}} = 0.8;{x_{{H_2}}} = 0.2\]

Answer 1

Hint: The sum of the partial pressures exerted by the component gases in a mixture is equal to the total pressure exerted by the mixture of the component gases.
Formula Used:
 \[{x_A} = \dfrac{{partial{\text{ }}pressure{\text{ }}of{\text{ A}}}}{{total{\text{ }}pressure{\text{ }}of{\text{ }}the{\text{ }}mixture}} = \dfrac{{{P_A}}}{{{P_T}}}\]

Complete step by step answer:
In the given reaction, a mixture of oxygen and hydrogen is ignited together. This reaction results in the formation of water. Now according to the conditions of this experiment, the water has been formed as result, is separated, and removed. After doing that, it is found that there are still traces of hydrogen left in the setup. The hydrogen that is left exerts a pressure of 0.40 atm at the same values of T and V as the original mixture. This ensures that the variation in pressure of hydrogen left is not because of the fluctuation in the values temperature and volume.
The reaction of this ignition can be represented as:
\[{x_A} = \dfrac{{partial{\text{ }}pressure{\text{ }}of{\text{ A}}}}{{total{\text{ }}pressure{\text{ }}of{\text{ }}the{\text{ }}mixture}} = \dfrac{{{P_A}}}{{{P_T}}}\]
This reaction can be balanced as follows:
\[2{H_2} + {O_2} \to 2{H_2}O\]
This shows that 3 moles of reactants are required to form 2 moles of product. The ratio of the number of moles of hydrogen to oxygen is 2:1.
The initial pressure of the mixture is 1.0 atm. After ignition of the mixture, the pressure of the hydrogen left is 0.40 atm. Hence, 0.60 atm of matter was consumed in the process. Hence, out of the 0.60 atm of pressure, the contribution of hydrogen is \[\dfrac{2}{3}\]and that of oxygen is \[\dfrac{1}{3}\] of the total consumed pressure.
Hence, 0.40 atm of oxygen and 0.2 atm of oxygen was consumed. and on top of that, another 0.40 atm of hydrogen is left.
Hence, in the initial setup of 1 atm pressure of the mixture, 0.2 atm was contributed by oxygen and the remaining (0.4 + 0.4 = 0.8 atm) is contributed by hydrogen. These pressure contributions by different gases to a mixture are also known as “Partial pressure of the constituent gas”.
Now, mole fraction of any constituent can be given by:
Mole fraction of A = \[{x_A} = \dfrac{{partial{\text{ }}pressure{\text{ }}of{\text{ }}A}}{{total{\text{ }}pressure{\text{ }}of{\text{ }}the{\text{ }}mixture}} = \dfrac{{{P_A}}}{{{P_T}}}\]
Hence,
\[{x_{{H_2}}} = \dfrac{{partial{\text{ }}pressure{\text{ }}of{\text{ }}{H_2}}}{{total{\text{ }}pressure{\text{ }}of{\text{ }}the{\text{ }}mixture}} = \dfrac{{{P_{{H_2}}}}}{{{P_T}}} = \dfrac{{0.8}}{{1.0}} = 0.8\]
\[{x_{{O_2}}} = \dfrac{{partial{\text{ }}pressure{\text{ }}of{\text{ }}{{\text{O}}_2}}}{{total{\text{ }}pressure{\text{ }}of{\text{ }}the{\text{ }}mixture}} = \dfrac{{{P_{{O_2}}}}}{{{P_T}}} = \dfrac{{0.2}}{{1.0}} = 0.2\]

Hence, Option A is the correct option.

Note:
Mole fraction is another way of expressing the concentration of a solution or mixture. It is equal to the moles of one component divided by the total moles in the solution or mixture. Mole fraction is used in a variety of calculations, but most notably for calculating partial pressures.