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 ?
-1368 kJ
To calculate the change in heat when a reaction occurs, you need to use the equation:
q = (moles of substance) x ∆H
First, we need to calculate the number of moles of Al in the solution.
Step 1: Calculate the number of moles of Al.
Given:
Volume of solution (V) = 0.455 L
Molarity of Al solution (M) = 3.60 M
We can use the formula:
moles of substance = (volume of solution) x (molarity of solution)
moles of Al = 0.455 L x 3.60 mol/L
moles of Al = 1.638 mol
Step 2: Calculate the change in heat.
Given:
∆H = +3339.6 kJ/mol (reaction enthalpy)
We will use the equation:
q = (moles of substance) x ∆H
q = 1.638 mol x 3339.6 kJ/mol
q = 5,465.2128 kJ
Therefore, the change in heat when 0.455 L of a 3.60 M Al solution reacts with excess O2 is approximately 5,465.2128 kJ.
To calculate the change in heat when a reaction occurs, you can use the equation:
∆H = q / n
where ∆H is the change in heat (in kJ/mol), q is the heat transferred (in kJ), and n is the number of moles of the reactant involved in the reaction.
First, we need to calculate the number of moles of Al that reacts. To do this, we can use the volume and concentration of the Al solution.
Step 1: Calculate the number of moles of Al using the volume and concentration.
Given:
Volume of Al solution = 0.455 L
Concentration of Al solution = 3.60 M
First, convert the volume from liters to moles using the molarity (M) of the solution:
moles of Al = volume (L) x concentration (M)
moles of Al = 0.455 L x 3.60 M
moles of Al = 1.638 mol
So, we have 1.638 moles of Al.
Step 2: Determine the number of moles of O2 required for the reaction.
According to the balanced chemical equation, the stoichiometry ratio of Al to O2 is 4:3. This means that for every 4 moles of Al that react, 3 moles of O2 are required.
Since we have 1.638 moles of Al, we can calculate the moles of O2 required using the stoichiometry ratio:
moles of O2 = (moles of Al) x (3 moles of O2 / 4 moles of Al)
moles of O2 = 1.638 mol x (3 mol O2 / 4 mol Al)
moles of O2 = 1.2245 mol
So, we need 1.2245 moles of O2 for the reaction.
Step 3: Calculate the change in heat (∆H).
Using the equation ∆H = q / n, we can rearrange the equation to solve for q:
q = ∆H x n
Given: ∆H = +3339.6 kJ/mol, n = 1.2245 mol
q = 3339.6 kJ/mol x 1.2245 mol
q = 4088.2 kJ
Therefore, the change in heat when 0.455 L of a 3.60 M Al solution reacts with excess O2 is 4088.2 kJ.