How much heat is evolved when 907 kg of Ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure) #N_2(g)# + #3H_2(g)# #rarr# #2NH_3(g)# ; #DeltaH# = -91.8 kJ Thank you so much! :)

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Answer 1 · 2018-02-28T21:36:39

#-2.31*10^6color(white)(l)kJ# of heat are released in the reaction.

Start by converting 907kg of #NH_3# to moles.

#(907000gcolor(white)(l)NH_3)/1*(1color(white)(l)molcolor(white)(l)NH_3)/(18.039gcolor(white)(l)NH_3)=5.03*10^4color(white)(l)molcolor(white)(l)NH_3#

Next, divide #DeltaH# by 2 to get it in terms of one mole of #NH_3#.

#-91.8/2=-45.9color(white)(l)kJ# per mole of #NH_3#

Finally, multiply#-45.9color(white)(l)kJ# by the number of moles ammonia.

#-45.9color(white)(l)kJ*(5.03*10^4color(white)(l)molcolor(white)(l)NH_3)=-2.31*10^6color(white)(l)kJ#