Consider the reaction

2 Al2O3(s) -> 4 Al(s) + 3 O2(g) .
�H = +3339.6 kJ/mol
How much heat is generated when 0.455 L
of a 3.60 M Al solution reacts with 1.05 mole
O2 ?

Well, let's calculate it step by step. First, we need to use the given molarity (3.60 M) and volume (0.455 L) of the Al solution to find the number of moles of Al.

So, moles of Al = Molarity x Volume
= 3.60 mol/L x 0.455 L
= 1.638 mol

Now, we can use the balanced equation to find the energy produced per mole of Al.

From the equation, we see that 4 moles of Al produces 3339.6 kJ.

Therefore, for 1 mole of Al, the energy produced = 3339.6 kJ / 4
= 834.9 kJ

Finally, we can find the total heat generated by multiplying the energy produced per mole of Al by the number of moles of Al.

Total heat generated = 834.9 kJ/mol x 1.638 mol
= 1365.2 kJ

So, the heat generated when 0.455 L of a 3.60 M Al solution reacts with 1.05 mole O2 is 1365.2 kJ. That's hot!

To calculate the heat generated in this reaction, we need to use the equation:

q = n * ΔH

Where:
q is the heat generated (in joules),
n is the number of moles of Al2O3 reacted (which we need to calculate),
ΔH is the enthalpy change of the reaction (+3339.6 kJ/mol).

First, let's calculate the number of moles of Al2O3 reacted using stoichiometry:

From the balanced equation, we can see that the stoichiometric ratio between Al2O3 and Al is 2:4, or 1:2. This means that 1 mole of Al2O3 produces 2 moles of Al.

Since we have 1.05 moles of O2, and according to the equation, 1 mole of O2 reacts with 2 moles of Al2O3, we can calculate the number of moles of Al2O3 as follows:

n(Al2O3) = 1.05 mol O2 * (2 mol Al2O3 / 1 mol O2) = 2.10 mol Al2O3

Now that we know the number of moles of Al2O3 (n = 2.10 mol), we can calculate the heat generated (q) using the equation:

q = n * ΔH

q = 2.10 mol * 3339.6 kJ/mol

q = 7014.36 kJ

Therefore, the heat generated when 0.455 L of a 3.60 M Al solution reacts with 1.05 moles of O2 is 7014.36 kJ.

Refer to the other heat problems.

Remember mols = M x L and that makes this a limiting reagent problem. You had some of those above but here is a link that will give you detailed instruction for solving limiting reagent problems.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

2736 J/K