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 ?
This is a duplicate post.
To calculate the change in heat for the given reaction, we need to use the concept of stoichiometry and the given reaction enthalpy.
1. Write out the balanced chemical equation:
2 Al2O3(s) → 4 Al(s) + 3 O2(g)
2. Determine the moles of Al in the reaction:
To calculate the moles of Al, we need the volume and concentration of the Al solution.
Given:
Volume of Al solution = 0.455 L
Concentration of Al solution = 3.60 M
Molarity (M) is defined as moles of solute per liter of solution. Since we have the concentration and volume, we can calculate the moles of Al:
Moles of Al = concentration (M) × volume (L)
Moles of Al = 3.60 M × 0.455 L
3. Calculate the heat change using the reaction enthalpy:
Given reaction enthalpy (ΔH) = +3339.6 kJ/mol
From the balanced equation, we see that 4 moles of Al are needed to produce a ΔH of +3339.6 kJ.
So we can set up a conversion:
4 moles Al = 3339.6 kJ
To find the heat change for the given number of moles of Al, we can set up a proportion:
(3339.6 kJ) / (4 moles) = X kJ / (moles of Al)
Cross-multiplying gives:
4 × X = (3339.6 kJ) × (moles of Al)
X = (3339.6 kJ × moles of Al) / 4
Now, substitute the value for moles of Al from step 2 into the equation and calculate X:
X = (3339.6 kJ × moles of Al) / 4
It is important to note that we need to convert moles of Al to moles of Al2O3, as the reaction enthalpy is given for the formation of 2 moles of Al2O3.
Since the molar ratio between Al and Al2O3 is 2:1, we have:
moles of Al2O3 = (moles of Al) / 2
Finally, substitute this value into the equation from step 3 to calculate the change in heat.