How many grams of water will be produced by the complete combustion of 10.00 g of octane (C8H18)?

Mass of water =
g H2O

To determine the mass of water produced by the complete combustion of octane (C8H18), we need to balance the chemical equation for the combustion of octane and then use stoichiometry to calculate the mass of water.

First, let's write the balanced chemical equation for the combustion of octane:

C8H18 + 12.5O2 → 8CO2 + 9H2O

From the balanced equation, we can see that for every 1 mole of octane combusted, we produce 9 moles of water.

Next, we need to calculate the number of moles of octane present in 10.00 g of octane. To do this, we use the molar mass of octane.

Molar mass of octane (C8H18):
8(12.01 g/mol) + 18(1.01 g/mol) = 114.22 g/mol

Now, we can calculate the number of moles of octane:
moles of octane = mass of octane / molar mass of octane
moles of octane = 10.00 g / 114.22 g/mol

Next, we use stoichiometry to find the number of moles of water produced:

moles of water = moles of octane * stoichiometric ratio of water
moles of water = moles of octane * 9

Finally, we can calculate the mass of water produced:

mass of water = moles of water * molar mass of water
mass of water = moles of water * (2(1.01 g/mol) + 16.00 g/mol)

Now, let's plug in the values and calculate the mass of water:

moles of octane = 10.00 g / 114.22 g/mol
moles of water = moles of octane * 9
mass of water = moles of water * (2(1.01 g/mol) + 16.00 g/mol)

mass of water = (10.00 g / 114.22 g/mol) * 9 * (2(1.01 g/mol) + 16.00 g/mol)

Evaluating this expression will give us the mass of water produced.