Heat produced when burning sugar C₁₂H₂₂O₁₁ in combustion reaction

[NekoVictoria]

This is what I was given:

Answer all of the questions concerning the combustion of table sugar,
$$C_1_2H_2_2O_1_1(s) + 12O_2 (g)\rightarrow 11H_2O(g) + 12CO_2(g)$$

a. If (weird triangle shape thing(what is this thing anyways?))H for this reaction is -5645 kj/mol, how much heat is produced when 2 mol of sugar are burned?

b. How much heat is produced when only 3 mol of $$O_2$$ are available

c. Could this combustion reaction ever have a positive (weird triangle shape)H? Explain.

I am completely stuck, I have no idea even where to start on this problem, and its due tomorrow.
A million thanks to anyone who can help me!

 

[symbolipoint]

(a)
Your reaction is written (although with a mistake) for 1 moles of the sugar compound. Your question is about 2 moles of this compound. The proportion should seem obvious.

(b)
First, balance the reaction correctly. The concept will then be similar to #(a), but this time, you are basing the decision on the oxygen.

 

[NekoVictoria]

That is not a mistake, that's exactly how the question is written.
How is it obvious? I don't see it at all... I have never had to do a question like this before now and it makes no sense, I have absolutely no idea how I'm supposed to calculate that...

 

[Borek]

That is not a mistake, that's exactly how the question is written.

There is an obvious typo in the reaction equation.

How is it obvious? I don't see it at all... I have never had to do a question like this before now and it makes no sense, I have absolutely no idea how I'm supposed to calculate that...

So you are in one of those schools where they want you to do questions not telling first what they are about... And they don't even ask you check your textbook... Sigh.

This strange triangle thing - Δ - is called delta.

ΔH is the amount of heat produced per mole of the reacting substance.

 

[rowkem]

I know this is probably useless but, I thought I would give it some closure since some replies were filled with snide remarks and little was done to resolve the question at hand.

First, I see nothing wrong with the equation, as far as I know, that's the combustion of sucrose and it's balanced correctly.

Either way, ΔH is directly proportional to the moles of, in this case, reactant(s). The concept will still apply if you're manipulating the moles of products. For your given ΔH, it only applies the combustion of sucrose in those specific quantities. Again, if its proportional to the moles:

A) You have twice as much sucrose now - so the combustion will produce twice as much heat: ΔH= 2(-5645 kj/mol)

B) You now have 1/4 of the oxygen you had originally - so combustion will produce only 1/4 of the heat: ΔH= (-5645 kj/mol) / 4

C) Combustion will NEVER be a positive ΔH. It's combustion, combustion always releases heat so ΔH will always be negative. Think about it; when was the last time a fire (essentially combustion) made the surroundings cold?

 

[Borek]

First, I see nothing wrong with the equation, as far as I know, that's the combustion of sucrose and it's balanced correctly.

Equation was edited, originally it contained 12O instead of 12O₂.