REVIEW 12/11, MOLES For the combustion reaction

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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. MOL OBJECTIVE 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 KNOWN STEP ONE:C4H8 to moles STEP TWO: C4H8 = 1 = 2.678 H20 4 X X = 10.71 MOL H20 STEP THREE: H2O to grams MOL = MASS GFW MOL = MASS GFW MOL = 150.0 = 2.678 MOL 56.0 C4H8 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 MOL OBJECTIVE 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 KNOWN STEP ONE: STEP TWO: C4H8 = 1 = 2.678 H20 4 X X = 10.71 MOL H20 STEP THREE: GAS VOL = MOL X 22.4 MOL = MASS GFW GAS VOL = 10.71 X 22.4 MOL = 150.0 = 2.678 MOL 56.0 C4H8 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. MOL OBJECTIVE 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 KNOWN STEP ONE: STEP TWO: C4H8 = 1 = 2.678 CO2 4 X X = 10.71 MOL CO2 STEP THREE: MOLARITY = MOL #L MOL = MASS GFW MOLARITY = 10.71 10.0L(aq) MOL = 150.0 = 2.678 MOL 56.0 C4H8 MOLARITY = 1.07 mol/L