source : yahoo.com
Combustion between Propane and Oxygen?
The belated picture shows 3 molecules of propane and 10 molecules of oxygen.
10 molecules of oxygen would react completely with 10 x (1/5) = 2 molecules of propane, but there is more propane present than that, so propane is in excess and O2 is the limiting reactant.
(10 molecules of O2) x (3 mol CO2 / 5 mol O2) = 6 molecules CO2
(10 molecules of O2) x (4 mol H2O / 5 mol O2) = 8 molecules H2O
3 molecules C3H8 initially – 2 molecules C3H8 reacted = 1 molecule C3H8 left over
No molecules of O2 will be left since O2 is the limiting reactant.
[I hope the comments aren’t important. YA won’t show them to me, so I haven’t seen them.]
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