Calculate the amount of heat evolved in the complete oxidation of #"8.17 g"# of #"Al"# #(M_M = "26.98 g/mol")# at #25^@"C"# and #"1 atm"# pressure?

The key here is the standard enthalpy change of formation, #DeltaH_f^@#, of aluminium oxide.

#DeltaH_f^@ = -"1676 kJ mol"^(-1)#

Note that the minus sign is used here to illustrate that the heat is being given off.

This value tells you that when #1# mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, #"1676 kJ"# of heat are being given off.

In other words, you know that standard enthalpy change for this reaction

#2"Al"_ ((s)) + 3/2 "O"_ (2(g)) -> "Al"_ 2"O"_( 3(s))#

is equal to

#DeltaH_f^@ = - "1676 kJ"#

This implies that this reaction

#(color(red)(2) * 2)"Al"_ ((s)) + (color(red)(2) * 3/2) "O"_ (2(g)) -> color(red)(2)"Al"_ 2"O"_( 3(s))#

which can be written as

#4"Al"_ ((s)) + 3"O"_ (2(g)) -> 2"Al"_ 2"O"_( 3(s))#

will have

#DeltaH_"rxn"^@ = color(red)(2) * DeltaH_f^@#

#DeltaH_"rxn"^@ = - "3352 kJ"#

So, when #2# moles of aluminium oxide are formed under standard conditions, the reaction gives off #"3352 kJ"# of heat.

Use the molar mass of aluminium to calculate the number of moles present in your sample

#8.17 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.30282 moles Al"#

According to the balanced chemical equation, the reaction will produce

#0.30282 color(red)(cancel(color(black)("moles Al"))) * ("2 moles Al"_2"O"_3)/(4color(red)(cancel(color(black)("moles Al")))) = "0.15141 moles Al"_2"O"_3#

You can thus say that when #"8.17 g"# of aluminium undergo complete combustion, the reaction will give off

#0.15141 color(red)(cancel(color(black)("moles Al"_2"O"_3))) * "3352 kJ kJ"/(2 color(red)(cancel(color(black)("moles Al"_2"O"_3)))) = color(darkgreen)(ul(color(black)("254 kJ")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of aluminium.

In other words, you have

#DeltaH_ ("rxn for 8.17 g Al")^@ = - "254 kJ"#

Once again, the minus sign is used to show that the heat is being given off.

Notice that you can get the same result by using the standard enthalpy change of formation of aluminium oxide

#0.15141 color(red)(cancel(color(black)("moles Al"_2"O"_3))) * "1676 kJ kJ"/(1 color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = color(darkgreen)(ul(color(black)("254 kJ")))#