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

it's filtration, but this is an assumption so my apologies if i got this