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! :)

1 Answer
Feb 28, 2018

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

Explanation:

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