Transcript STOICIOMETRY9_14_07.ppt

3.64) GIVEN 1 Gallon of C8H18 ( D=0.692 g./ mL.), find grams of O2
needed to react completely.
(
1 Gallon
)( ) ( )
3785.4 mL
1 Gallon
Research, 1 gallon =
3.7854 L = 3785.4
mL
2 C8H18
INITIAL
22.97mol
CHANGE
-2X
0.692 g.
1 mol
mL
114.0 g.
= 22. 97mol
C8H18
2619.22 g
+ 25 O2
16 CO2 + 18 H2O
287.125mol
0
0
- 25 X
+ 16 X
+ 18 X
NOTE: 2 X = 22.970for the known0C8H18, therefore
X in all
terms is
END
16(11.48)
18(11.48)
11.48. Always solve for x relative to limiting reagent!
C8H18
O2
=
2
25
=
22.97mol X = 287.125mol O2
X
287. 125 Mol O x 32 g/mol = 9184 g
2
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.