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 ?
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
1 answer
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