How do you determine the amount of heat (in kJ) given off when #1.26 xx 10^4# g of ammonia are produced according to the equation #N_2(g) + 3H_2(g) -> 2NH_3(g)#, #DeltaH°_(rxn)# = -92.6 kJ/mol?

The problem provides you with the thermochemical equation that describes the synthesis of ammonia from nitrogen gas and hydrogen gas.

#"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))" "DeltaH_ "rxn"^@ = - "92.6 kJ mol"^(-1)#

Notice that the standard enthalpy change of reaction, #DeltaH_"rxn"^@#, is equal to #-"92.6 kJ mol"^(-1)#. This tells you that when #1# mole of ammonia is produced by the reaction, #"92.6 kJ"# of heat are being given off.

Keep in mind that the minus sign attached to the standard enthalpy change of reaction signifies heat lost.

#DeltaH_"rxn" = color(red)(-)"92.6 kJ"color(white)(.)color(blue)("mol"^(-1)) => "92.6 kJ"color(white)(.)color(red)("lost")color(white)(.)color(blue)("per mole")color(white)(.)"of NH"_3# #"produced"#

Your first step will be to convert the number of grams of ammonia to moles by using the compound's molar mass

#1.26 * 10^4 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.031color(red)(cancel(color(black)("g")))) = "739.83 moles NH"_3#

So, if #"92.6 kJ"# of heat are being given off when #1# mole is produced, it follows that when #739.83# moles are being produced, the reaction will give off

#739.83 color(red)(cancel(color(black)("moles NH"_3))) * "92.6 kJ"/(1color(red)(cancel(color(black)("mole NH"_3)))) = color(darkgreen)(ul(color(black)("68,500 kJ")))#

The answer is rounded to three sig figs.

Keep in mind that when #"68,500 kJ"# of heat are being given off, the standard enthalpy change of reaction must carry a minus sign!

#DeltaH_ ("rxn for 739.83 moles NH"_ 3)^@ = - "68,500 kJ"#