Please Help Me!! I am begging you! I can't figure out this Chemistry problem for the life of me!?

Question

Oxidation of gaseous CIF by F2 yields liquid ClF3 an important flourinating agent. Use the following thermochemical equations to calculate the standard heat of reaction for the production of chlorine trifluoride (ClF3).
1) 2ClF(g) + O2 => Cl2O(g ) + OF2 (g) change in H = 167.5KJ
2) 2F2(g) + O2(g) => 2OF2(g) change in H = -43.5KJ
3) 2CLF3(l) + 2O2(g) => Cl2O(g) + 3OF2(g)
Based on your answer, how much heat is released (positive number) when 4.99 grams of ClF react with excess fluorine and oxygen gas to form ClF3 under standard conditions? Express your answer in kJ. Anyone know the answer? Please Help!

Answer 1

So, the reaction you're looking at is:

ClF + F2 --> ClF3

Start with the first equation as it is written, and its DH (delta H)

2 ClF + O2 --> Cl2O + OF2

Flip the third equation and change the sign of its DH

3 OF2 + Cl2O --> 2 O2 + 2 ClF3

Add the second equation as written with its DH
2F2 + O2 --> 2 OF2

Adding those three equations gives you:
2 ClF + 2 F2 --> 2 ClF3

Adding the DHs for the three reactions (with their correct signs) gives you DH for that equation.

Finally, divide that last equation AND its DH by 2 to get the equation you are after and its DH.

That DH is for 1 mole of ClF reacting. Just convert your 4.99 grams ClF into moles and multiply your DH by that and you're done.

Answer 2

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Answer 3

note:whatever change you make in a reaction must be applied even on delta H
1.multiply the first reaction by 1/2 or divide it by 2. do the same thing in delta h
2.multiply the second reaction by 1/2 or divide by 2. also in delta H
3.for reaction 3. exchange the reactant and product.so that the reactant are at RHS and product at LHS. then also change the sign for delta H. THEN MULTIPLY IT BY 1/2.or divide by 2. do it also in delta H
4.chancel the same reactant to other side.thus same reactant to left chancel with one to right.
5. add delta H values..in this case it will be 135.05