Checkpoints Questions Help

[b^3]

Ok guys I'm having a bit of trouble understanding how checkpoints got their answer to these questions.

Q37d)For each fuel calculate the mass of carbon diozide formed when 1.00 kJ of energy is released.
There are two fuels, ethanol and octane. Here is the other releveant information from the previous parts of the questions.



Q38) Copper forms two oxides, Cu₂O and CuO. The energy change when each oxide is formed is given by the equation below.


Use this information to calculate the for the reaction.

I am just having a little bit of trouble putting the two equations together. I know you have to reverse and half the second one, but cannot see how it fits into the other.

Thanks guys & girls.

[luken93]

37; for every 2 mols of CO2 produced, it produces 1364kj
 = 1 mol produces 682kJ.
1/682 = 0.001467 mols
etc.

38)
Reverse:   2 Cu2O  ->  4 Cu  +  O2;   338kJ

Add the first reaction to get:
2 Cu  +  O2  +  2 Cu2O ->   2 CuO  +  4 Cu  +  O2   (-310.4 + 338)
=  2 Cu2O ->   2 CuO  +  2 Cu  27.6 kJ
Cu2O  ->  CuO  +  Cu   13.8 kJ

[b^3]

37; for every 2 mols of CO2 produced, it produces 1364kj
 = 1 mol produces 682kJ.
1/682 = 0.001467 mols
etc.

38)
Reverse:   2 Cu2O  ->  4 Cu  +  O2;   338kJ

Add the first reaction to get:
2 Cu  +  O2  +  2 Cu2O ->   2 CuO  +  4 Cu  +  O2   (-310.4 + 338)
=  2 Cu2O ->   2 CuO  +  2 Cu  27.6 kJ
Cu2O  ->  CuO  +  Cu   13.8 kJ

Thanks for 38, for 37 can you show the working because thats the way I looked at it and didn't get the right answer.

[b^3]

answers are for octane: 0.0644g of CO₂
ethanol: 0.0645g of CO₂

[luken93]

From what I had written, 1/682 mol = 1kJ of energy.
m(CO2) = 1/682 * 44 = 0.0645

For Octane,
for every 16 mols of CO2 produced, it produces 5464 kJ
 = 1 mol produces 341.5 kJ.
1/341.5 * 44 = 0.129g?

Hmmm, not sure, I'll have a look in a sec, I think I brought them home....

[Mao]

Q37d)For each fuel calculate the mass of carbon diozide formed when 1.00 kJ of energy is released.
There are two fuels, ethanol and octane. Here is the other releveant information from the previous parts of the questions.

You are interpreting the questions incorrectly. Firstly, these are exothermic reactions, thus should be negative. Secondly, the heat of combustion (presumably from the data booklet) are per mole of fuel. I.e. 5464 kJ of heat per mole of octane. thus, the chemical equation should be

[Kiro Kompiro]

For 37 I'd do it this way:

For ethanol: 1 mol of ethanol burned releases 1364 kJ.  So 1 kJ must mean 1/1364 mol of ethanol burned.
 From the balanced equation n(ethanol)=n(C02)/2.  Therefore n(CO2)=2 x n(ethanol)=2 x 1/1364 mol
M(CO2)=44, so mass of CO2=44 grams/mol x 2 x 1/1364 mol=0.0645 grams of CO2.

For Octane, 1 kJ is release so 1/5464 mol of octane must have burned.
From the balanced equation n(octane)/2=n(Co2)/16
ie n(CO2)= 16 x n(octane)/2 = 16 x (1/5464)/2= 0.001464 mol
M(CO2) is 44 so m(octane) is 44 x 0.001464=0.0644 grams.

If anyone is interested

My dad showed me quick maths way of dealing with mass-mol calcs:

Balanced equation: aA + bB----> cC + dD, then

n(A)           n(B)        n(C)        n(D)
----      =   ------   = -------  =  -------
  a               b             c             d

[b^3]

Q37d)For each fuel calculate the mass of carbon diozide formed when 1.00 kJ of energy is released.
There are two fuels, ethanol and octane. Here is the other releveant information from the previous parts of the questions.

You are interpreting the questions incorrectly. Firstly, these are exothermic reactions, thus should be negative. Secondly, the heat of combustion (presumably from the data booklet) are per mole of fuel. I.e. 5464 kJ of heat per mole of octane. thus, the chemical equation should be

Thanks Mao and guys. And yeh I know it's exothermic, I wrote it with negative in my book, forgot it when I typed it up, originally I tried doing it using 1 mol to 5464kJmol^-1 and it didn't work, must of punched the wrong values in the calculator. Anyway thanks guys