how many grams of oxygen are required to react with 18.0 grams of octane (C8H18) in the combustion of octane in gasoline?
I already balanced the equation just need to know the answer. Really bad at figuring it out
- pisgahchemistLv 71 month ago
Really bad? If you learn how to use the unit-factor method of problem solving (aka dimensional analysis) then you will be a whiz at doing stoichiometry problems. It will be nearly impossible to make a mistake.
2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g)
18.0g ............. ???g
18.0g C8H18 x (1 mol C8H18 / 114.0g C8H18) x (25 mol O2 / 2 mol C8H18) x (32.0g O2 / 1 mol O2) = 63.2g O2
But the actual answer is nearly ZERO. That's because there is essentially no octane (C8H18) in gasoline. It comprises less than 1% of the mixture we call gasoline. Just because there is an "octane rating" for gasoline doesn't mean that gasoline contains octane.
The "octane rating" doesn't even use n-octane. It uses 2,2,4-trimethyl pentane (C8H18), called "iso-octane", and n-heptane. The rating is the percentage of 2,2,4-trimethylpentane in a mixture of it and heptane that performs like the test gasoline.
- Anonymous1 month ago
Start with the balanced equation.