Gaseous arsine reacts according to the balanced equation. If a total of -3663 KJ of heat is produced in the reaction as written, how many grams of AsH3 are consumed? (Enthalpy= -1867.2)

Question

2 AsH3(g) + 3 O2(g) --> As2O3(s) + 3 H2O(l)

Answer 1

3663 kJ of heat to produce AsH3.
Delta H, according to value in parentheses is 1867.2 kJ/mol I assume, then you must have had 3363/1867.2 = 1.801 mols.
1.8 mol x molar mass = grams.

Answer 2

while you were doing the calculations, i think u mistyped the numbers...

i did your steps though..

i divided -3663/-1867.2 = 1.9618 mols

then i multiply 1.9618 mols x 77.946 = 152.91 grams

but then answer is wrong..the correct answer is 306

what did i do wrong?

Answer 3

You're right. I did make a typo in the math part but I think your answer is correct. Take a close look at your problem. If the problem states 3663 kJ of heat for the REACTION AS WRITTEN (and not 3663 kJ/mol) and if the delta H you have in parentheses is 1867.2 kJ/MOL), then 3663/1867.2 = 1.962 mols and that time 77.94 = 152.9 grams AsH3. You can see that 152.9 x 2 = 305.8 which rounds to 306 but the 2 coefficient for AsH3 has already been taken into account if the problem you have is as stated in your post. Therefore, I think the answer sheet/computer program/whateveryouhave is not right. There is a common problem, sometimes, in these thermochemical problems, to leave the per mol out or to quote only a per mol basis and expect us to know to multiply by the coefficient. I don't think that's the case here. Good luck.

Answer 4

I see it as this:
Heat of reaction is based on Kj/moleproduct
moles of product As2O3=1.9618mols
Moles of Reactant AsH3=2*1.9618
grams of AsH3=2*1.9618*molmassAsH3
=(74.93+3.022)2*1.96=306grams