Question #857fa

2 Answers
Dec 17, 2014

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

Dec 17, 2014

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#.