review 12/11, 2007 -- moles

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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. 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 :C 4 H 8 to moles MOL = MASS GFW MOL = 150.0 = 2.678 MOL 56.0 C 4 H 8 STEP TWO : C 4 H 8 = 1 = 2.678 H 2 0 4 X X = 10.71 MOL H 2 0 STEP THREE : H 2 O to grams MOL = MASS GFW 10.71 = X = 192.81 g 18.0 H 2 O C+ 4 x 12.0 = 48.0 H 8 x 1.00 = + 8.00 56.0 G/MOL

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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 Presentation

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Page 1: REVIEW 12/11, 2007 -- MOLES

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

Page 2: 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 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

Page 3: 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 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