[Parothwk20]

I have 48.7 g of ZnS burning in the presence of O giving me 220 KJ of heat.  I am trying to find the DH for the reaction====== 2ZnS  +  3O2   ---->  2ZnO   +  2SO2

I am not sure how to begin.  My book provides zero examples and very limited literature on how to approach this; my teacher only uses H20 and and heat as examples in class providing little exposure on how to do this.  I have tried wikipedia but it is far to complicated to follow.  Any Ideas??  Thanks for the time.

[enahs]

What is the units of ΔH?

How much energy did the reaction provide, in how many mols?

[Parothwk20]

the units of ΔH should be in KJ

and I think there is 0.25 mol of ZnS?

I do not know how to find the amount of energy the reaction provided, I think this is my main problem?

[enahs]

I do not know how to find the amount of energy the reaction provided, I think this is my main problem?

That information is provided in the question.

[Parothwk20]

From what I understand

2ZnS + 3O2 + 220KJ => 2ZnO + 2SO2

(48.7 g ZnS)(1mol/194.8 g) = 0.25 mol ZnS

(220KJ)/(48.7g) = 4.52 KJ/g

(4.52 KJ/g)(194.8 g) = 882KJ

ΔH = 882KJ

Does this appear correct?

[Parothwk20]

I have also tried it this way

220KJ = (48.7 g ZnS)(2mol ZnS/194.8g)(ΔH/1 mol ZnS)

Solve for ΔH = 441 KJ

[Parothwk20]

I have tried a few other problems similiar to this one, I think 441 KJ is the correct answer, does anyone else agree? 

[Borek]

There is 220 kJ per 0.25 mole of ZnS. Reaction uses 8 times that amount of ZnS (2/0.25), so the correct answer seems to be 8*220. Something like

220 kJ = (ΔH/2 mol ZnS)(48.7 g ZnS)(1 mol ZnS/194.8 g ZnS)

[Parothwk20]

That does not match up with the HW solution, it is a multiple choice question and 1760 KJ is not an answer.

[Sev]

That's because 48.7g is not 0.25 mol.

[Borek]

That's because 48.7g is not 0.25 mol.

Yep. I have blindly copied 194.8 g as molar mass of ZnS, but that is wrong value.