1. ( ) ( ) ( ) 2 C8H18 + 25 O2 16 CO2 + 18 H2O INITIAL 287.125mol CHANGE
3.64) GIVEN 1 Gallon of C8H18 ( D=0.692 g./ mL.), find grams of O2 needed to react completely.
1 Gallon mL g mol = mol
1 Gallon mL g C8H18 ( ) ( ) ( )
Research, 1 gallon = L = mL g
2 C8H18
+ 25 O2
16 CO2
+ 18 H2O
INITIAL
mol
CHANGE
- 2 X
- 25 X
+ 16 X
+ 18 X
END
16(11.48)
18(11.48)
22.97mol
NOTE: 2 X = for the known C8H18, therefore X in all terms is Always solve for x relative to limiting reagent!

2. 2 X = 287.125mol O2 25 X 287. 125 Mol O2 x 32 g/mol = 9184 g C8H18
= =
C8H18 2
22.97mol
X = mol O2 25 O2 X
Mol O2 x 32 g/mol = 9184 g
NOTE♫
In this calculation the density has the least significant figures (3), therefore the answer should round to 3. The line over the eight indicates it is the last significant digit. Remember round only the answer, not the intermediate numbers.