Ideal Gas Equilibrium Question

In an ideal gas mixture, the following reaction is studied:

2 N2O(g) + N2H4(g) ⇌ 3 N2(g) + 2 H2O(g)

Initially there are 0.10 moles of N2O and 0.25 moles of N2H4 in a 10.0-L container (with
no N2 or H2O initially present). If there are 0.04 moles of N2O at equilibrium, how many
moles of N2 are present at equilibrium?

The answer is 0.09. But I don't get it. Below is my logic.
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I clearly don't understand the problem, because I think it's a lot easier than it actually must be and I'm getting the wrong answer. If there's a 3:2 N2 to N2O mol ratio, and there are 0.04 mol N2O at equilibrium, shouldn't there just be 0.04 * (3/2) mol N2 at equilibrium? 0.04 * 1.5 = 0.06 mol N2. But the answer says it's 0.09 mol. I don't get what else I'm supposed to do, some sort of ideal gas law equation?

Initially there were 0.10 moles of N2O,
and they used up 0.04 moles of N2O by the time they got to equilibrium,
so that means that 0.06 moles of N2O reacted

by the equation:
2 N2O(g) + N2H4(g) ⇌ 3 N2(g) + 2 H2O(g)
that 0.06 moles of N2O which reacted, produces 3/2's as many moles of N2 = 0.09 mol N2