Question

Question #857fa

Answer 1

The answer is `1.53` moles of `Al`.

First' let's start with the balanced chemical equation

`Fe_2O_3(s) + 2Al_((s)) -> 2Fe_((s)) + Al_2O_3(s)`

Notice that we have a `1:2` mole ratio between `Fe_2O_3` and `Al`, and a `1:1` mole ratio between `Fe_2O_3` and `Al_2O_3`, since this will become useful later on.

Here are the standard state enthalpy value for the reactans and the products

`Fe_2O_3` - `DeltaH_f^@` = `-822.2 (kJ)/(mol e)`;
`Al` - `DeltaH_f^@` = `0 (KJ)/(mol e)`;
`Al_2O_3` - `DeltaH_f^@` = `-1669.8 (kJ)/(mol e)`;
`Fe` - `DeltaH_f^@ = 0 (kJ)/(mol e)`;

For this reaction. `DeltaH` is equal to the sum of the `DeltaH_f^@`'s of the products minus the sum of the `DeltaH_f^@`'s of the reactants - each mutiplied by their stoichiometric coefficients

`DeltaH` = `(-1669.8 (kJ)/(mol e)) * 1 mol e + (0 (kJ)/(mol e) * 2 mol es) - ((-822.2 (kJ)/(mol e) * 1 mol e) + 0 (kJ)/(mol e) * 2 mol es)) = -850 kJ`

Since this reaction is exothermic, the heat given off will be equal to

`q = -DeltaH = 850kJ`

However, the heat given off is set to be `q = 650kJ`, less than what we've calculated so far; this means that `DeltaH` is bigger (since -650 is bigger than -850), which in turns means that fewer moles reacted.

Since the moles of `Al` and `Fe` do not influence the reaction's enthalpy, we'll focus on `Fe_2O_3`. Let's assume we have x moles of `Fe_2O_3`, instead of 1 mole, to start with. `DeltaH` will become

`DeltaH` = `(-1669.8 (kJ)/(mol e)) * x. mol es + (0 (kJ)/(mol e) * 2x. mol es) - ((-822.2 (kJ)/(mol e) * x. mol es) + 0 (kJ)/(mol e) * 2x. mol es)) = -650 kJ`

So `-1669.8 * x + 822.2 * x = -650 -> x = 0.767`

So `0.767` moles of `Fe_2O_3` are used, which means that the number of moles of `Al` is

`n_(Al) = 2 * n_(Fe) = 2 * 0.767 = 1.53` moles

Here's a video of the reaction - the famous thermite

Answer 2

1.53 moles are reacted.

`2Al+Fe_2O_3rarrAl_2O_3+2Fe`

`DeltaH=-850kJ`

So 850 kJ is produced from 2 mol Al

1kJ produced from 2/850 = 0.00235 mol

So

650 kJ produced from 0.00235 x 650 = 1.53 mol

If have assumed `DeltaH` is per mole of `Fe_2O_3`.