Transcript 439REVIEW_12_11_07_

REVIEW 12/11, 2007 -- MOLES
For the combustion reaction
C4H8 (g) + 6 O2 (g)  4 CO2 (g) + 4 H2O (g)
a) If 150.0 g. of C4H8 (g) reacts, calculate the mass of H2O that would be produced.
STEP ONE:
IDENTIFY YOUR
KNOWN AND
CONVERT TO
MOLES
MOL KNOWN
STEP ONE:C4H8 to moles
MOL = MASS
GFW
4 x
H
8 x
STEP TWO:
C4H8 = 1 = 2.678
H20
4
X
MOL = 150.0 = 2.678 MOL
56.0
C4H8
C+
STEP TWO :MOLE
(COEFFICIENT)
RATIO OF KNOWN
TO OBJECTIVE
12.0 =
1.00 =
X = 10.71 MOL H20
48.0
+ 8.00
56.0 G/MOL
MOL
OBJECTIVE
STEP THREE :
CONVERT
OBJECTIVE TO
UNITS
STEP THREE: H2O to grams
MOL = MASS
GFW
10.71 = X =
18.0
192.81 g
H2O
For the combustion reaction
C4H8 (g) + 6 O2 (g)  4 CO2 (g) + 4 H2O (g)
a) If 150.0 g. of C4H8 (g) reacts, calculate the STP VOLUME of H2O that would be
produced.71
STEP ONE:
IDENTIFY YOUR
KNOWN AND
CONVERT TO
MOLES
MOL KNOWN
STEP ONE:
MOL = MASS
GFW
MOL = 150.0 = 2.678 MOL
56.0
C4H8
STEP TWO :MOLE
(COEFFICIENT)
RATIO OF KNOWN
TO OBJECTIVE
STEP TWO:
C4H8 = 1 = 2.678
H20
4
X
MOL
OBJECTIVE
STEP THREE :
CONVERT
OBJECTIVE TO
UNITS
STEP THREE:
GAS VOL = MOL X 22.4
GAS VOL = 10.71 X 22.4
X = 10.71 MOL H20
GAS VOL = 239.90 L
For the combustion reaction
C4H8 (g) + 6 O2 (g)  4 CO2 (g) + 4 H2O (g)
a) If 150.0 g. of C4H8 (g) reacts, calculate the MOLARITY of of CO2 (aq) that would
be produced, assume a solution volume of 10.0L.
STEP ONE:
IDENTIFY YOUR
KNOWN AND
CONVERT TO
MOLES
MOL KNOWN
STEP ONE:
MOL = MASS
GFW
MOL = 150.0 = 2.678 MOL
56.0
C4H8
STEP TWO :MOLE
(COEFFICIENT)
RATIO OF KNOWN
TO OBJECTIVE
STEP TWO:
C4H8 = 1 = 2.678
CO2
4
X
MOL
OBJECTIVE
STEP THREE :
CONVERT
OBJECTIVE TO
UNITS
STEP THREE:
MOLARITY = MOL
#L
MOLARITY = 10.71
X = 10.71 MOL CO2
10.0L(aq)
MOLARITY = 1.07 mol/L