The following reaction is investigated (assume an ideal gas mixture): 2N20(g) + N2H4(g) = 3N2(g) + 2H20(g) Initially there are 0.100 mol of N20 and 0.25 mol of N2H4, in a 10.0-L container. If there are 0.062 mol of N20 at equilibrium, how many moles of N2 are present at equilibrium? 3.8 x 10-2 O1.1 x 10-1 5.7 x 10-2 1.9 x 10-2 none of these

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