2A + N2 -> 2AIN How many moles of aluminum nitride form when 5.00 g of aluminum reacts with excess nitrogen? K 0.0925 mol 0 0.185 mol Oleo mol z se mol​

H tricky one let me think and get back to u

Ibelieve the answer is b.2200 watts, because p=vi; p=220(10); then p=2200hope my answer has you : )

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.