My question my question

H2O as an oxidant only

Given

Reaction

2 Na + 2 H2O → 2 NaOH + H2

Required

The function of water in the equation

Solution

Water : oxidizing agent

Na : reducing agent

Na⁰ → Na⁺ (oxidation)

H⁺- → H⁰ (reduction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H⁺ ion)

Base = proton (receiver) acceptor (H⁺ ion)

If water is acting as an acid it should only give 1 H +, so that becomes:

H₂O (l) ⇔ OH⁻ (aq) not H₂

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.