Consider the reaction
2 Al2O3(s) → 4 Al(s) + 3 O2(g).
∆H = +3339.6 kJ/mol
What is the change in heat when 0.455 L of
a 3.60 M Al solution reacts with excess O2 ?
1. -21,962 kJ
2. -2735 kJ
3. -1368 kJ
4. -26,521 kJ
5. -685 kJ
6. -5477 kJ
7. -424 kJ
8. -1695 kJ
Technically, I don't think this is a viable problem because it gives a SOLUTION of Al while the rxn is Al solid.
mols Al = M x L = approx 1.7 but you need to do this more accurately.
Then -3339.6 kJ/4 mol Al x 1.7 mol Al = approx ?
To find the change in heat when the reaction occurs, we can use the formula:
ΔH = (moles of Al2O3) * ΔH(molar) / volume(L)
First, let's calculate the moles of Al2O3 present in the solution:
Molarity (M) is defined as moles of solute per liter of solution. Given that the solution has a concentration of 3.60 M and a volume of 0.455 L, we can calculate the moles of Al in the solution as follows:
moles of Al = Molarity * Volume
= 3.60 mol/L * 0.455 L
≈ 1.638 mol
Since the balanced equation shows that 2 moles of Al2O3 produce 4 moles of Al, we can conclude that 1.638 moles of Al2O3 will produce (2/4) * 1.638 = 0.819 moles of Al.
Now we can calculate the change in heat using the given value of ΔH:
ΔH = (moles of Al2O3) * ΔH(molar) / volume
= 0.819 mol * (+3339.6 kJ/mol) / 0.455 L
≈ 5979.152 kJ
Round the answer to three significant figures:
ΔH ≈ 5979 kJ
Looking at the given options, none of them match the calculated value exactly. However, the closest option is -5477 kJ (option 6).