100 sq cm of water gas is exploded with 100sq cm of oxygen and the residual gases are cooled to room temperature . Find the volume composition of the resulting mixture.
100 sq cm of water gas is exploded with 100sq cm of oxygen and the residual gases are cooled to room temperature . Find the volume composition of the resulting mixture.
100 cc each of water gas and oxygen are ignited and theresultant mixture og gases cooled to room temperature. Calculate the composition of the resultant mixture . The combustion reaction of water gas can be written as: CO + H2 + O2→ CO2 + H2O (l)
First, we have to find the limiting reagent upon which the concentration of carbon dioxide depends.
Volume of water gas = 100 cc
Volume of CO = 50 cc = 0.05 L and Volume of H2 = 50 cc = 0.05 L ( as they are present in equal ratio in water gas).
Volume of O2 = 100 cc = 0.1 L
According to balanced chemical equation,
CO + H2 + O2→ CO2 + H2O(l)
1 mole or 22.4 L of O2 reacts with 1 mole or 22.4 L of CO.
So, 0.1 L of O2 will react with (22.4 x 0.1) / 22.4 = 0.1 L of CO.
As the given amount of CO is less than the required amount therefore, CO is the limiting reagent.
Now, 22.4 L of CO forms 22.4 L of CO2.
So, 0.05 L of CO will form (0.05x22.4) / 22.4 = 0.05 L of CO2.
Now, the amount of CO2 formed = 0.05 L = 50 cc
The amount of O2 left unreacted = 100 cc - 50 cc = 50 cc
Again, by balanced chemical equation,
22.4 L of H2 reacts with 22.4 L of O2.
Thus, 0.05 L of H2 will react with (0.05x22.4) / 22.4 = 0.05 L of O2.
Therefore, the remaining 50 cc of O2 will react with 50cc of H2 to from water.
The resultant mixture will contain only 50 cc CO2.
CO + H2 + O2→ CO2 + H2O (l)
First, we have to find the limiting reagent upon which the concentration of carbon dioxide depends.
Volume of water gas = 100 cc
Volume of CO = 50 cc = 0.05 L and Volume of H2 = 50 cc = 0.05 L ( as they are present in equal ratio in water gas).
Volume of O2 = 100 cc = 0.1 L
According to balanced chemical equation,
CO + H2 + O2→ CO2 + H2O(l)
1 mole or 22.4 L of O2 reacts with 1 mole or 22.4 L of CO.
So, 0.1 L of O2 will react with (22.4 x 0.1) / 22.4 = 0.1 L of CO.
As the given amount of CO is less than the required amount therefore, CO is the limiting reagent.
Now, 22.4 L of CO forms 22.4 L of CO2.
So, 0.05 L of CO will form (0.05x22.4) / 22.4 = 0.05 L of CO2.
Now, the amount of CO2 formed = 0.05 L = 50 cc
The amount of O2 left unreacted = 100 cc - 50 cc = 50 cc
Again, by balanced chemical equation,
22.4 L of H2 reacts with 22.4 L of O2.
Thus, 0.05 L of H2 will react with (0.05x22.4) / 22.4 = 0.05 L of O2.
Therefore, the remaining 50 cc of O2 will react with 50cc of H2 to from water.
The resultant mixture will contain only 50 cc CO2.
