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)
2 AsH3(g) + 3 O2(g) --> As2O3(s) + 3 H2O(l)
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.
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?
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.
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
To determine the number of grams of AsH3 consumed in the reaction, we need to use the given enthalpy change and the balanced equation.
First, let's convert the enthalpy change from kJ to J by multiplying by 1000:
-3663 kJ * 1000 J/kJ = -3663000 J
Next, we can use the following equation to relate the enthalpy change to the amount of substance consumed or produced:
ΔH = n * ΔH°
Where:
ΔH is the enthalpy change (-3663000 J)
n is the number of moles of AsH3
ΔH° is the standard enthalpy change (-1867.2 J/mol)
Rearranging the equation, we can solve for n:
n = ΔH / ΔH°
n = -3663000 J / -1867.2 J/mol
n ≈ 1963.44 mol
From the balanced equation, we know that 2 moles of AsH3 react to produce 1 mole of As2O3. Therefore, the number of moles of As2O3 produced is half the number of moles of AsH3 consumed:
n(As2O3) = 1963.44 mol / 2 = 981.72 mol
Finally, we can use the molar mass of AsH3 (Molar mass AsH3 = 74.92 g/mol) to find the number of grams of AsH3 consumed:
Mass AsH3 = n(AsH3) * Molar mass AsH3
Mass AsH3 = 1963.44 mol * 74.92 g/mol ≈ 146942.09 g
Therefore, approximately 146942.09 grams of AsH3 are consumed in the reaction.