24cm^3 of a mixture of methane and ethane were exploded with 90cm^3 of oxygen. After cooling to RTP the volume of gases was noted and it decreased by 32cm^3 when treated with KOH. Calculate the composition of the mixture.
CH4 + 2O2 ==> CO2 + 2H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O
I would look at this. Since we have all gases we need not convert to mols but we can use volume directly.
Let x = volume CH4
and y = volume C2H6
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x + y = 24 cc which is equation 1.
x cc of CH4 will produce x cc of CO2.
y cc of C2H6 will produce 4y/2 = 2y CO2 so equation 2 is
x + 2y = 32. Solve those two equations simultaneously for cc x and cc y.
Thank you DrBobb22
To calculate the composition of the mixture, we can use the concept of stoichiometry and the ideal gas law.
First, let's break down the problem and understand the information given:
1. Initial volume of the mixture = 24 cm^3
2. Volume of oxygen used = 90 cm^3
3. Volume decrease after cooling and treatment with KOH = 32 cm^3
Now, let's start solving the problem step by step:
Step 1: Determine the volume of gases produced through the reaction between the mixture and oxygen.
According to the reaction equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
The stoichiometric ratio between methane and oxygen is 1:2. Therefore, for every unit of methane reacted, we need two units of oxygen. Since we used 90 cm^3 of oxygen, the volume of methane in the mixture must be half of that, so 90 cm^3 / 2 = 45 cm^3.
Step 2: Determine the volume decrease after cooling and treatment with KOH.
When the resulting gases from the reaction are treated with KOH, any carbon dioxide that is present will be absorbed. The volume decrease after treatment is due to the carbon dioxide being removed.
Given that the volume decreased by 32 cm^3, we can deduce that the volume of carbon dioxide in the mixture is 32 cm^3.
Step 3: Calculate the volume of remaining gases, which is the sum of methane and ethane, after the removal of carbon dioxide.
The remaining volume of gases after the absorption of carbon dioxide is the initial volume minus the volume decrease, which is 24 cm^3 - 32 cm^3 = -8 cm^3.
Since volume cannot be a negative value, it means that all of the carbon dioxide was absorbed, and the initial volume of 24 cm^3 was entirely comprised of methane and ethane. Therefore, the composition of the mixture is 100% methane and 0% ethane.
In conclusion, the mixture consists of 100% methane and 0% ethane.