The combustion of octane produces carbon dioxide and water. If you start with 10.0 mol of octane, what mass of water (in grams) would you expect to produce?

2C8H18 + 25 O2 ==> 16CO2 + 18H2O

10 mols Octane x (18 mols H2O/2 mol octane) = 10 x 18/2 = 90 mols H2O

g H2O = mols x molar mass = ? if 100% yield

To find the mass of water produced, we need to use stoichiometry.

The balanced equation for the combustion of octane is:
2C8H18 + 25O2 → 16CO2 + 18H2O

From this equation, we can see that 2 moles of octane produce 18 moles of water.

Given that we start with 10.0 mol of octane, we can set up a simple ratio to find the moles of water produced:

10.0 mol octane × (18 mol water / 2 mol octane) = 90 mol water

Now, to find the mass of water produced, we need to use the molar mass of water. The molar mass of water (H2O) is approximately 18.02 g/mol.

Mass of water produced = 90 mol water × 18.02 g/mol = 1,621.8 g

Therefore, you would expect to produce 1,621.8 grams of water.