C15H30 is a hydrocarbon that reacts with oxygen gas to give carbon dioxide and water.

Calculate the mass of water (grams) formed when 724.05 grams of the hydrocarbon reacts with the oxygen gas.

This is a stoichiometry problem. Here is a worked example. Just follow the steps. You need not convert grams to moles since you are given moles initially.

http://www.jiskha.com/science/chemistry/stoichiometry.html

C15H30 +02 -> CO2 + H2O

Im having trouble balancing it.
Thank you

2C15H30 + 45O2 ==> 30CO2 + 30H2O

To calculate the mass of water formed when the hydrocarbon reacts with oxygen gas, we need to determine the balanced chemical equation for the reaction.

The balanced equation for the combustion of hydrocarbons can be represented as follows:

C15H30 + (23/2)O2 -> 15CO2 + 15H2O

From the equation, we can see that for every mole of C15H30, we get 15 moles of H2O produced.

To determine the amount of moles of C15H30, we need to convert the given mass of the hydrocarbon (724.05 grams) to moles using the molar mass of C15H30.

The molar mass of C15H30 is calculated by summing up the atomic masses of all the atoms in it.

For carbon (C), the atomic mass is 12.011 g/mol, and for hydrogen (H), the atomic mass is 1.008 g/mol.

So the molar mass of C15H30 can be calculated as follows:

(15 * 12.011 g/mol) + (30 * 1.008 g/mol) = 450.682 g/mol

Now, we can calculate the amount of moles of C15H30:

Moles = Mass / Molar mass
Moles = 724.05 g / 450.682 g/mol ≈ 1.6079 mol

Since the molar ratio between C15H30 and H2O is 1:15, we can calculate the amount of moles of water produced:

Moles of H2O = 1.6079 mol * 15 = 24.1185 mol

Finally, we can calculate the mass of water formed by multiplying the moles of H2O by its molar mass. The molar mass of H2O is (2 * 1.008 g/mol) + (16.00 g/mol) = 18.016 g/mol.

Mass of water = Moles of H2O * Molar mass of H2O
Mass of water = 24.1185 mol * 18.016 g/mol ≈ 434.53 grams

Therefore, approximately 434.53 grams of water are formed when 724.05 grams of the hydrocarbon react with the oxygen gas.