Question

When a chemist burns ammonia according to the reaction below she finds that the reaction releases heat. (It is exothermic.)
4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)
The enthalpy of the reaction DH = -1267 kJ. What is the enthalpy change (in kJ) when 7 grams of ammonia are burned?
You need to find the arrow key and use it. Without an arrow how do we tell the reactants from the products. Fortunately I know where it is but in the future remember this tip.

dHrxn is -1267 kJ and that is for 4*17 = 68g NH3. YOu want to know dH for 7 g NH3.

That's -1267 kJ x (7/68) = ? kJ

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