# Question 857fa

Dec 17, 2014

The answer is $1.53$ moles of $A l$.

$F {e}_{2} {O}_{3} \left(s\right) + 2 A {l}_{\left(s\right)} \to 2 F {e}_{\left(s\right)} + A {l}_{2} {O}_{3} \left(s\right)$

Notice that we have a $1 : 2$ mole ratio between $F {e}_{2} {O}_{3}$ and $A l$, and a $1 : 1$ mole ratio between $F {e}_{2} {O}_{3}$ and $A {l}_{2} {O}_{3}$, since this will become useful later on.

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

$F {e}_{2} {O}_{3}$ - $\Delta {H}_{f}^{\circ}$ = $- 822.2 \frac{k J}{m o l e}$;
$A l$ - $\Delta {H}_{f}^{\circ}$ = $0 \frac{K J}{m o l e}$;
$A {l}_{2} {O}_{3}$ - $\Delta {H}_{f}^{\circ}$ = $- 1669.8 \frac{k J}{m o l e}$;
$F e$ - $\Delta {H}_{f}^{\circ} = 0 \frac{k J}{m o l e}$;

For this reaction. $\Delta H$ is equal to the sum of the $\Delta {H}_{f}^{\circ}$'s of the products minus the sum of the $\Delta {H}_{f}^{\circ}$'s of the reactants - each mutiplied by their stoichiometric coefficients

$\Delta H$ = (-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

$q = - \Delta H = 850 k J$

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

Since the moles of $A l$ and $F e$ do not influence the reaction's enthalpy, we'll focus on $F {e}_{2} {O}_{3}$. Let's assume we have x moles of $F {e}_{2} {O}_{3}$, instead of 1 mole, to start with. $\Delta H$ will become

$\Delta H$ = (-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 $- 1669.8 \cdot x + 822.2 \cdot x = - 650 \to x = 0.767$

So $0.767$ moles of $F {e}_{2} {O}_{3}$ are used, which means that the number of moles of $A l$ is

${n}_{A l} = 2 \cdot {n}_{F e} = 2 \cdot 0.767 = 1.53$ moles

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

Dec 17, 2014

1.53 moles are reacted.

$2 A l + F {e}_{2} {O}_{3} \rightarrow A {l}_{2} {O}_{3} + 2 F e$

$\Delta H = - 850 k J$

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 $\Delta H$ is per mole of $F {e}_{2} {O}_{3}$.