Question #857fa

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Question #857fa

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Answer 1 · 2014-12-17T12:46:17

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

First' let's start with the balanced chemical equation

$F e_{2} O_{3} (s) + 2 A l_{(s)} \to 2 F e_{(s)} + A l_{2} O_{3} (s)$ $Fe_2O_3(s) + 2Al_((s)) -> 2Fe_((s)) + Al_2O_3(s)$

Notice that we have a $1 : 2$ $1:2$ mole ratio between $F e_{2} O_{3}$ $Fe_2O_3$ and $A l$ $Al$ , and a $1 : 1$ $1:1$ mole ratio between $F e_{2} O_{3}$ $Fe_2O_3$ and $A l_{2} O_{3}$ $Al_2O_3$ , since this will become useful later on.

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

$F e_{2} O_{3}$ $Fe_2O_3$ - $Δ H_{f}^{\circ}$ $DeltaH_f^@$ = $- 822.2 \frac{k J}{m o l e}$ $-822.2 (kJ)/(mol e)$ ;
$A l$ $Al$ - $Δ H_{f}^{\circ}$ $DeltaH_f^@$ = $0 \frac{K J}{m o l e}$ $0 (KJ)/(mol e)$ ;
$A l_{2} O_{3}$ $Al_2O_3$ - $Δ H_{f}^{\circ}$ $DeltaH_f^@$ = $- 1669.8 \frac{k J}{m o l e}$ $-1669.8 (kJ)/(mol e)$ ;
$F e$ $Fe$ - $Δ H_{f}^{\circ} = 0 \frac{k J}{m o l e}$ $DeltaH_f^@ = 0 (kJ)/(mol e)$ ;

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

$Δ H$ $DeltaH$ = $(- 1669.8 \frac{k J}{m o l e}) \cdot 1 m o l e + (0 \frac{k J}{m o l e} \cdot 2 m o l e s) - ((- 822.2 \frac{k J}{m o l e} \cdot 1 m o l e) + 0 \frac{k J}{m o l e} \cdot 2 m o l e s)) = - 850 k J$ $(-1669.8 (kJ)/(mol e)) * 1 mol e + (0 (kJ)/(mol e) * 2 mol es) - ((-822.2 (kJ)/(mol e) * 1 mol e) + 0 (kJ)/(mol e) * 2 mol es)) = -850 kJ$

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

$q = - Δ H = 850 k J$ $q = -DeltaH = 850kJ$

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

Since the moles of $A l$ $Al$ and $F e$ $Fe$ do not influence the reaction's enthalpy, we'll focus on $F e_{2} O_{3}$ $Fe_2O_3$ . Let's assume we have x moles of $F e_{2} O_{3}$ $Fe_2O_3$ , instead of 1 mole, to start with. $Δ H$ $DeltaH$ will become

$Δ H$ $DeltaH$ = $(- 1669.8 \frac{k J}{m o l e}) \cdot x . m o l e s + (0 \frac{k J}{m o l e} \cdot 2 x . m o l e s) - ((- 822.2 \frac{k J}{m o l e} \cdot x . m o l e s) + 0 \frac{k J}{m o l e} \cdot 2 x . m o l e s)) = - 650 k J$ $(-1669.8 (kJ)/(mol e)) * x. mol es + (0 (kJ)/(mol e) * 2x. mol es) - ((-822.2 (kJ)/(mol e) * x. mol es) + 0 (kJ)/(mol e) * 2x. mol es)) = -650 kJ$

So $- 1669.8 \cdot x + 822.2 \cdot x = - 650 \to x = 0.767$ $-1669.8 * x + 822.2 * x = -650 -> x = 0.767$

So $0.767$ $0.767$ moles of $F e_{2} O_{3}$ $Fe_2O_3$ are used, which means that the number of moles of $A l$ $Al$ is

$n_{A l} = 2 \cdot n_{F e} = 2 \cdot 0.767 = 1.53$ $n_(Al) = 2 * n_(Fe) = 2 * 0.767 = 1.53$ moles

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

Answer 2 · 2014-12-17T12:52:13

1.53 moles are reacted.

$2 A l + F e_{2} O_{3} \to A l_{2} O_{3} + 2 F e$ $2Al+Fe_2O_3rarrAl_2O_3+2Fe$

$Δ H = - 850 k J$ $DeltaH=-850kJ$

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 $Δ H$ $DeltaH$ is per mole of $F e_{2} O_{3}$ $Fe_2O_3$ .

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