C15H30 is a hydrocarbon that reacts with oxygen gas to give carbon dioxide and water.
Calculate the mass of water (grams) formed when 859.64 grams of the hydrocarbon reacts with the oxygen gas.
This is a problem in stoichiometry. Here is a link to a problem I've posted that is an example. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the mass of water formed when 859.64 grams of the hydrocarbon reacts, we first need to balance the chemical equation for the reaction:
C15H30 + O2 -> CO2 + H2O
Next, we need to determine the molar mass of C15H30. C15H30 is comprised of 15 carbon atoms and 30 hydrogen atoms. The atomic mass of carbon is 12.01 g/mol, and the atomic mass of hydrogen is 1.01 g/mol.
The molar mass of C15H30 can be calculated as follows:
(15 carbon atoms * 12.01 g/mol) + (30 hydrogen atoms * 1.01 g/mol) = 450.45 g/mol
Now, we need to use the balanced chemical equation to determine the stoichiometry of the reaction. From the balanced equation, we can see that for every mole of C15H30 that reacts, we get 15 moles of water.
The number of moles of C15H30 can be calculated by dividing the given mass by the molar mass:
859.64 g / 450.45 g/mol = 1.908 moles of C15H30
Using the stoichiometry of the reaction, we can determine the number of moles and the mass of water formed using the following ratio:
1.908 moles of C15H30 * 15 moles of H2O / 1 mole of C15H30 = 28.62 moles of H2O
Finally, we can calculate the mass of water formed by multiplying the number of moles by the molar mass of water:
28.62 moles of H2O * 18.02 g/mol = 516.25 grams of H2O
Therefore, when 859.64 grams of C15H30 reacts with oxygen gas, the mass of water formed is 516.25 grams.