[insins88]

For the below calculation I got an answer of -854kj but the answer given is -1699kj. Am I right?



Methanol (CH3OH), which is used as a cooking fuel, undergoes combustion to produce carbon dioxide and water:

2CH3OH(g) + 3O2(g) --> 2CO2(g) + 4H2O(l)ΔH = -726 kJ

How many kilojoules are released when 75.0 g methanol is burned? -

1699 kJ (this is the answer given)

[ARGOS++]

Dear insins88;

IMHO:  I thought that ∆H_Combustion is always given per mole, not per reaction equation.

Good Luck!
                    ARGOS⁺⁺

[Borek]

ΔH = -726 kJ

kJ or kJ/mol? If just kJ, you are right. If kJ/mol, you are wrong.

In fact it is about -730 kJ/mol - so your result is wrong. But if the question was worded exactly as you have posted it, your approach was correct, just the question was lousy.

[insins88]

Yeah, this is only CHM130.

My working:

75.0 g CH3OH x 1 mol CH3OH  x  -726 kJ  = -851 kJ
                      32.0 g CH3OH       2 mol CH3OH

[ARGOS++]

Dear insins88;

NO!,  -  If I use -726 kJ/mole, then I get -1699.33 kJ/75.0g.
Don’t divide by 2 moles of CH₃OH.

Good Luck!
                    ARGOS⁺⁺

[insins88]

I just copied and pasted this from my lesson:

Calculations using ΔH
The value of ΔH tells us the heat change for the number of moles of each substance in the balanced equation.

For example,

2H2O ® 2H2 + O2         ΔH = 572 kJ

The above gives use three usable conversion factors:

572 kJ   572 kJ   572 kJ
2 mol H2O 2 mol H2 1 mol O2

We can use those conversion factors to convert between moles and kJ.

Ex: 2CH3OH + 3O2 ® 2CO2 + 4H2O ΔH = -726 kJ
How many kJ are released when 75.0 g of CH3OH is burned?

Step 1: You know you need to be calculating in moles so you can convert to kJ, so you first need to convert grams into moles (using molar mass).

75.0 g CH3OH x  1 mol CH3OH
32.0 g CH3OH


Step 2: Now that you are calculating in moles of CH3OH, you are able to convert to kJ.

75.0 g CH3OH x  1 mol CH3OH  x  -726 kJ
32.0 g CH3OH 2 mol CH3OH


Step 3: Multiply the numbers on top and divide the numbers on bottom.

75.0 g CH3OH x  1 mol CH3OH  x  -726 kJ  = -851 kJ
32.0 g CH3OH 2 mol CH3OH


Which to me seems like I have been given the same question twice but each has a different answer.

[ARGOS++]

Dear insins88;

I can’t really correct read your copy from the lesson.

But 726 kJ/mole is the ∆H_Combustion for Methanol, according to:
          http://en.wikipedia.org/wiki/Heat_of_combustion#Heat_of_combustion_tables 

So I believe more in this reference.
Good Luck!
                    ARGOS⁺⁺