A gaseous mixture having 1 lit ethane and 1 lit propane was completely burnt. What volume of O2 will
be required? (pls give a solution as well! I will also mark your solution as brainliest only if the solution is given.)
Answers
Explanation:
Assuming STP though it really doesn't matter since gases have volumes relative to their molar weight.
Getting your head around it.
At STP we have.
1 mole =22,400 cm3. (22.4 liters)
Of course all gases are affected by temperature or pressure and change equally.
But at STP we have for the methane or ethylene for that matter
50 cm3/22,400 cm3 = 1/448 moles
= 0.002232142857143 moles
Of course this value changes with temperature and pressure. Since you were asked for volume that doesn't matter. It doesn't matter because when you measure volumes of gases they all change together and the molar ratios don't change.
Methane molar mass
CH4 → 12 + 4 * 1 = 16
At STP the mass would be.
16/448= 1/28 grams = 0.035714285714286 grams
Deal only with volumes.
Since you weren't given temperature or pressure or mass so you don't know precisely how much.
Regardless you know the precise ratios!! Which is the point.
The reactions
Methane
C + O2 → CO2
12 + 2 * 16 = 44 grams/mol
4 H + O2 → 2 H2O
Which per mol is
4 * 1 + 2 * 16 = 2 * 18 = 36 grams/mol
Two moles of oxygen gas are consumed by one mole of methane and since and we started with 50 cm3 we consume 100 cm3 of oxygen at the same temperature and pressure by the methane alone. Whatever mass it turns out to be. (though at STP it would be 2*32/448=1/7 gram.
Ethylene
C2H4 by inspection we know
2 C + 2 O2 → 2 CO2 two moles of oxygen
4 H + O2 → 2 H2O one mole of oxygen.
So 3 moles of Oxygen are consumed by each mole of ethylene. And since we started with 50 cm3 of ethylene we needed 150 cm3 of oxygen for the ethylene alone.
Total
100 + 150 = 250 cm3 Oxygen.
At STP 5*32/448 = 5/14 gram.
Though it's not needed for this problem it's always good to have an idea of what's going on.
Answer:
Explanation:
combustion of Hydrocarbon given carbon dioxide and water
For ethane;
C₂H₆+7/2O₂ → 2CO₂ + 3H₂O [Balanced equation]
Here, every 1 mole of ethane required 7/2 moles of oxygen.
Now, For propane;
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Here, 1 mole of propane required 5 moles of oxygen.
As we know, 1 mole of any gas has volume of 22.4 L at NTP
so, 1 L of ethane and propane both have 1/22.4 mole each in the mixture
so, for complete comustion of Ethane, Required moles of O₂ is 7/2 *1/22.4 = 0.156 moles
And, for complete comustion of Propane, Required moles of O₂ is 5 * 1/22.4 = 0.223moles
Thus, Total Required moles of oxygen is 0.223mol. + 0.156 mol. = 0.388moles
Now, 1 mole corresponds to 22.4 L
Hence, 0.388 moles of oxygen have 0.388 * 22.4 L = 8.69 Liters
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