Asked by Devin
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?
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?
Answers
Answered by
DrBob222
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
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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