Calculating Delta H of A_2 Reaction with B_2

[Dooh]

Ok, I am stuck half way through this problem. here's what i got:

a hyopthetical substance A_2 reacts with B_2:

3 A_2 + 4 B_2 --> 2 A_3 B_4

delta H = ? kcal/mole A_2

1 mole of A = 56.4g
1 mole of B = 29.6g

when 15.6g of A_2 and an excess of B_2 react in a calorimeter that contains 0.00186 mL of water, the water temp changes from 23/5 celsius to 86.7 celsius. What is the delta H of the reaction in kcal/mole?

I approached this by first solving for q , then divide it by mole. But I am not sure if i solved for the mole correctly because i multiplied 56.4g by 6 since its "3 A_2". Also, after i get the answer, am i suppose to divide it by 3 because in the reaction, the coefficient for A_2 was 3, that doesn't match with what the answer wanted, which is kcal/ mole A_2. (coefficient is 1) Any help?

 

[chem_tr]

Hi, did you notice that A is bimolecular, as in A₂? In addition, you are right to include its coefficient in calorimetric calculation, since 3 moles of A₂ reacts with an excess of B₂. One mole of A₂ is equal to 2*56.4=112.8 grams. Be careful...

 

[GCT]

You know that $q_{reaction} = -(q_{solution} + q_{calorimeter})$. $q_{solution}$refers to water and you should know what $q_{calorimeter}$represents. In the end simply divide $q_{reaction}$ by the molar quantity of the limiting reagent, than use factor labeling using the coefficient ratio of $\frac{moles lim reagent}{moles rxn}$.