2.28 g of Octane is completely burned in 7.00 g of Oxygen...CHEMISTRY HELP!?
2.28 g of Octane is completely burned in 7.00 g of Oxygen...HELP!?
1) Which is the limiting reagent?
2) How much water vapor can be produced?
- DougLv 51 decade agoFavorite Answer
Well complete combustion of a hydrocarbon such as octane reacts with oxygen to produce only carbon dioxide and water.
C8H18 + O2 -> CO2 + H2O
To solve this problem we need to balance this chemical equation first:
C8H18 + ?O2 -> 8CO2 + 9H2O (note I have already balanced carbon and hydrogen here, each side of the equation has 8 carbon atoms and 18 hydrogen atoms)
Now looking at the right side of the equation, there are 16 + 9 = 25 oxygen atoms. So the appropriate coefficient for O2 is 25/2. So to combust one mole of C8H18 completely, one needs 25/2 moles of O2
So now, we need to find the amount of moles of C8H18 and O2 available.
Approximate Molecular weights: C = 12 H = 1 O = 16 (if you need more precise weights, please refer to a periodic table)
C8H18 = 8(12) + 18(1) = 114 g/mol
O2 = 2(16) = 32 g/mol
moles of C8H18 = 2.28 g C8H18(1 mol C8H18/ 114 g C8H18) = 0.020 mol C8H18
moles of O2 = 7.00 g O2(1 mol O2/ 32 g O2) = 0.21875 mol O2
the amount of oxygen needed to completely combust all of the octane:
(0.02 mol C8H18)(25 mol O2 / 2 mol C8H18) = 0.25 mol O2
Since we only have 0.21875 mol of O2 available, the oxygen will run out before completely combusting all of the octane. This makes Oxygen the limiting reagent.
b) Since we know that O2 is the limiting reagent, all of the oxygen will be consumed and then the reaction will come to a halt. From our balanced chemical equation, for every 25/2 (12.5) moles of O2 there are 9 moles of H2O produced.
So the amount of water produced from the reaction is:
0.21875 mol O2(9 mol H2O/ 12.5 mol O2) = 0.1575 mol H2O
One can use the molecular weight of water (approximately 18 g/mol) to find the mass of water vapor produced, which is: 2.385 grams.