2) What is the change in enthalpy when 3.400 moles of aluminum react with excess ammonium nitrate (NH4NO3) according to the equation? Is the reaction exothermic or endothermic?

2AI + 3NHANO3 → 3N2+ 6 H2O + Al2O3
AH = -2030kJ

See below

Explanation:

(i) 2AI + 3 NHANO3 → 3 N2+ 6 H2O + Al2O3  ΔH = -2030kJ

From stoichiometric coefiicients, it can be seen that:

When 2 moles of Al react, change in enthalpy =  -2030kJ

So, when 3.4 moles of Al react, change in enthalpy =  -(2030/2) x 3.4  kJ = -3451 kJ

Therefore, the change in enthalpy is -3451 kJ

(ii) The change in enthalpy is negative. This means heat is released during the reaction.

So the reaction is exothermic

The first one, or A

Explanation:

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