What volume of oxygen gas at STP would be needed for the complete combustion of 39.0 grams of octane ?
where did you get the 12.5
there are 25 O2 so there are 12.5 O
Why did the octane cross the road? To find some oxygen gas for a combustion party at STP, of course!
Now, let me crunch some numbers for you. The molar mass of octane (C8H18) is approximately 114.22 g/mol. So, if you have 39.0 grams of octane, that corresponds to approximately 0.341 moles.
Now, for the combustion reaction of octane:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
We can see that for every 2 moles of octane, we need 25 moles of oxygen gas.
Using stoichiometry, we can find out how much oxygen gas is needed for 0.341 moles of octane:
0.341 moles octane * (25 moles O2 / 2 moles octane) = 4.26 moles of oxygen gas
Now let's convert that to volume at STP (Standard Temperature and Pressure).
1 mole of any gas occupies 22.4 liters at STP. So,
4.26 moles O2 * (22.4 liters O2 / 1 mole O2) = 95.46 liters of oxygen gas at STP.
Voila! Approximately 95.46 liters of oxygen gas at STP would be needed for the complete combustion of 39.0 grams of octane.
To find the volume of oxygen gas needed for the complete combustion of octane, we need to follow a few steps:
Step 1: Calculate the moles of octane.
To do this, we need to know the molar mass of octane, which is C8H18. We can find the molar mass by summing the atomic masses of all the elements in the compound.
Molar mass of octane:
(12.01 g/mol * 8) + (1.01 g/mol * 18) = 114.23 g/mol
To find the moles, we divide the mass of octane by its molar mass:
39.0 g / 114.23 g/mol = 0.341 mol
Step 2: Determine the balanced chemical equation.
The balanced equation for the combustion of octane is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
From the balanced equation, we can see that 2 moles of octane react with 25 moles of oxygen gas.
Step 3: Calculate the moles of oxygen gas needed.
Using the stoichiometry of the balanced equation, we can determine the moles of oxygen gas required to react completely with the moles of octane.
0.341 mol octane * (25 mol O2 / 2 mol octane) = 4.2625 mol O2
Step 4: Convert moles to volume at STP.
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, we can convert the moles of oxygen gas to volume at STP.
4.2625 mol O2 * 22.4 L/mol = 95.35 L
So, the volume of oxygen gas needed for the complete combustion of 39.0 grams of octane at STP is approximately 95.35 liters.
22.4 L/mole ... gas at stp
octane is 114 g/mole
2 C8H18(g) + 25 O2(g) ---> 16 CO2 (g) + 18 H2O(l)
(39.0 / 114) * 12.5 * 22.4 L