review 12/11, 2007 -- moles
DESCRIPTION
REVIEW 12/11, 2007 -- MOLES. For the combustion reaction C 4 H 8 (g) + 6 O 2 (g) 4 CO 2 (g) + 4 H 2 O (g) a) If 150.0 g. of C 4 H 8 (g) reacts, calculate the mass of H 2 O that would be produced. MOL OBJECTIVE. STEP ONE : IDENTIFY YOUR KNOWN AND CONVERT TO MOLES. - PowerPoint PPT PresentationTRANSCRIPT
REVIEW 12/11, 2007 -- MOLESFor 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
STEP TWO :MOLE (COEFFICIENT) RATIO OF KNOWN TO OBJECTIVE
STEP THREE : CONVERT OBJECTIVE TO UNITS
MOL OBJECTIVE
MOL KNOWN
STEP ONE:C4H8 to moles
MOL = MASS GFW
MOL = 150.0 = 2.678 MOL 56.0 C4H8
STEP TWO:
C4H8 = 1 = 2.678H20 4 X
X = 10.71 MOL H20
STEP THREE: H2O to grams
MOL = MASS GFW
10.71 = X = 192.81 g 18.0 H2O
C+ 4 x 12.0 = 48.0
H 8 x 1.00 = + 8.00
56.0 G/MOL
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
STEP TWO :MOLE (COEFFICIENT) RATIO OF KNOWN TO OBJECTIVE
STEP THREE : CONVERT OBJECTIVE TO UNITS
MOL OBJECTIVE
MOL KNOWN
STEP ONE: STEP TWO:
C4H8 = 1 = 2.678H20 4 X
X = 10.71 MOL H20
STEP THREE:
MOL = MASS GFW
MOL = 150.0 = 2.678 MOL 56.0 C4H8
GAS VOL = MOL X 22.4
GAS VOL = 10.71 X 22.4
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
STEP TWO :MOLE (COEFFICIENT) RATIO OF KNOWN TO OBJECTIVE
STEP THREE : CONVERT OBJECTIVE TO UNITS
STEP ONE: STEP TWO:
C4H8 = 1 = 2.678CO2 4 X
X = 10.71 MOL CO2
STEP THREE:
MOL = MASS GFW
MOL = 150.0 = 2.678 MOL 56.0 C4H8
MOLARITY = MOL #L
MOLARITY = 10.71
10.0L(aq)
MOLARITY = 1.07 mol/L
MOL OBJECTIVE
MOL KNOWN