The combustion of propane and oxygen produces carbon dioxide and water. What mass of water is produced from the combustion of 10.4g of propane, C3H8

C3H8 + 5O2 ==> 3CO2 + 4H2O

mols C3H8 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols C3H8 to mols H2O
Now convert mols H2O to g H2O. g = mols x molar mass.

To determine the mass of water produced from the combustion of 10.4g of propane (C3H8), we need to know the balanced chemical equation for the combustion reaction.

The balanced chemical equation for the combustion of propane is as follows:
C3H8 + 5O2 → 3CO2 + 4H2O

From this equation, we can see that for every 1 mole of propane (C3H8) combusted, 4 moles of water (H2O) are produced.

First, we need to convert the mass of propane into moles. To do this, we divide the mass of propane by its molar mass.

The molar mass of propane (C3H8) can be calculated as:
(3 * molar mass of carbon) + (8 * molar mass of hydrogen)

The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.01 g/mol.

Therefore, the molar mass of propane (C3H8) is:
(3 * 12.01 g/mol) + (8 * 1.01 g/mol) = 44.11 g/mol

Now, we can calculate the number of moles of propane in 10.4g by dividing the mass by the molar mass:
10.4 g / 44.11 g/mol = 0.236 mol

According to the balanced chemical equation, 1 mole of propane produces 4 moles of water. Therefore, we can determine the number of moles of water produced by multiplying the number of moles of propane by the stoichiometric ratio:
0.236 mol propane * 4 mol water / 1 mol propane = 0.944 mol water

Finally, we can convert the moles of water to grams by multiplying by the molar mass of water (H2O), which is approximately 18.015 g/mol:
0.944 mol water * 18.015 g/mol = 16.9 g

Therefore, the mass of water produced from the combustion of 10.4g of propane is approximately 16.9 grams.