
Given,100 cc each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temperature. Calculate the composition of the resultant mixture. [ Water gas contains CO and ${H_2}$in equal ratio].
Answer
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Hint: We have been given that the water gas is a mixture of equal amount of CO and ${H_2}$. So, their total amount will be equal to the volume of water gas. Thus, the amount of Carbon monoxide and hydrogen can be easily found out. From the following reaction, we can see the number of moles reacting and then the amount of resultant mixture can be easily calculated.
$CO(g) + {H_2}(g) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l)$
Complete answer:
Let us write what is given to us and what we need to find out.
Given :
Volume of water gas = 100 cc
Volume of oxygen = 100 cc
To find :
Composition of resultant mixture
We have also been given that water gas consists of carbon monoxide and hydrogen gas.
Thus, the reaction that takes place is as -
$CO(g) + {H_2}(g) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l)$
We have Volume of water gas = volume of CO + Volume of ${H_2}$= 100 cc
So, Volume of CO = 50 cc
Volume of ${H_2}$= 50 cc
From Gay-Lussac’s law,
We can say that 1 mole of CO needs 1 mole of oxygen.
So, 50 mol of CO will react with = 50 cc of oxygen
Further, from the above reaction, we see that 1 mol of CO produce = 1 mole of Carbon dioxide
So, 50 cc of CO = 50 cc of carbon dioxide
Therefore, the composition of resultant mixture after the reaction is as-
Amount of CO = Amount present - Amount used
Amount of CO = 50 - 50
Amount of CO = 0
Amount of Hydrogen = Amount present - Amount used
Amount of Hydrogen = 50 - 50
Amount of Hydrogen = 0
Amount of water = Negligible
Amount of Carbon dioxide = 50
Amount of Oxygen = Amount present - Amount used
Amount of Oxygen = 100 - 50
Amount of Oxygen = 50
Note:
It must be noted that Gay Lussac’s law states that the pressure of a given mass of gas is directly proportional to absolute temperature of gas, keeping the volume constant. It is also called Pressure-temperature law.
$CO(g) + {H_2}(g) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l)$
Complete answer:
Let us write what is given to us and what we need to find out.
Given :
Volume of water gas = 100 cc
Volume of oxygen = 100 cc
To find :
Composition of resultant mixture
We have also been given that water gas consists of carbon monoxide and hydrogen gas.
Thus, the reaction that takes place is as -
$CO(g) + {H_2}(g) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l)$
We have Volume of water gas = volume of CO + Volume of ${H_2}$= 100 cc
So, Volume of CO = 50 cc
Volume of ${H_2}$= 50 cc
From Gay-Lussac’s law,
We can say that 1 mole of CO needs 1 mole of oxygen.
So, 50 mol of CO will react with = 50 cc of oxygen
Further, from the above reaction, we see that 1 mol of CO produce = 1 mole of Carbon dioxide
So, 50 cc of CO = 50 cc of carbon dioxide
Therefore, the composition of resultant mixture after the reaction is as-
Amount of CO = Amount present - Amount used
Amount of CO = 50 - 50
Amount of CO = 0
Amount of Hydrogen = Amount present - Amount used
Amount of Hydrogen = 50 - 50
Amount of Hydrogen = 0
Amount of water = Negligible
Amount of Carbon dioxide = 50
Amount of Oxygen = Amount present - Amount used
Amount of Oxygen = 100 - 50
Amount of Oxygen = 50
Note:
It must be noted that Gay Lussac’s law states that the pressure of a given mass of gas is directly proportional to absolute temperature of gas, keeping the volume constant. It is also called Pressure-temperature law.
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