If I have 2.5L of propane at STP, how many grams of water can be produced by this combustion?
Ok so this is the equation:
C3H8 + 5O2 = 3CO2 + 4H2O
What do I have to do next?
So you have 0.1116 mols propane. According to your balanced equation, 1 mol C3H8 will produce 4 mols H2O. That means 0.1116 mols C3H8 will produce .......mols H2O.
I know STP is 22.4 L, but I need more help
so, you have 2.5/22.4 = 0.1116 moles of propane
Now look at the reaction equation for the combustion to see how many moles of water that will produce
then convert that to grams
Ahh, I get it now thank you
To determine the number of grams of water produced by the combustion of 2.5 liters of propane at standard temperature and pressure (STP), we need to follow a few steps:
Step 1: Convert 2.5 liters of propane to moles.
Propane's molar volume (at STP) is approximately 22.4 liters/mol. Therefore, we can calculate the number of moles of propane by dividing the volume (2.5L) by the molar volume (22.4L/mol).
Moles of propane = 2.5L / 22.4L/mol
Step 2: Use the balanced chemical equation.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
This equation tells us that for every mole of propane combusted, we obtain 4 moles of water.
Step 3: Calculate moles of water produced.
Using the mole ratio from the balanced equation, we can determine the number of moles of water produced by multiplying the moles of propane calculated in Step 1 by the mole ratio.
Moles of water = Moles of propane × (4 moles of water / 1 mole of propane)
Step 4: Convert moles of water to grams.
To convert moles of water to grams, we need to use the molar mass of water, which is approximately 18 g/mol.
Grams of water = Moles of water × Molar mass of water
By following these steps, you can determine the number of grams of water produced by the combustion of 2.5 liters of propane at STP.