Asked by Beth
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. -685 kJ
2. -21,962 kJ
3. -1368 kJ
4. -2738 kJ
5. -5477 kJ
6. -26,521 kJ
7. -424 kJ
8. -1695 kJ
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. -685 kJ
2. -21,962 kJ
3. -1368 kJ
4. -2738 kJ
5. -5477 kJ
6. -26,521 kJ
7. -424 kJ
8. -1695 kJ
Answers
Answered by
kyle
Molarity = moles/Liters so moles = Molarity * Liters
moles = 3.60 M * .455 L = 1.64 moles of Al
1.64 moles Al / 4 moles Al = .4095 (this number is how many times the reaction can be done) .4095 moles * -3339.6 kj/mol (negative because you're doing reverse reaction and reverse reaction is reversed heat) = -1368 kj
answer = -1368 kj
moles = 3.60 M * .455 L = 1.64 moles of Al
1.64 moles Al / 4 moles Al = .4095 (this number is how many times the reaction can be done) .4095 moles * -3339.6 kj/mol (negative because you're doing reverse reaction and reverse reaction is reversed heat) = -1368 kj
answer = -1368 kj
Answered by
z
-2738
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