Question

| 11. The following reaction is investigated (assume an ideal gas mixture). 2N2O(g) + N2H4(g) 3N2(g) + 2H2O(g) Initially there are 0.08 mol of NO and 0.35 mol of N,H4, in a 30.0-L container. If there is 0.050'mol of N20 at equilibrium, how many moles of N2 are present at equilibrium? A) 0.15 B) 0.030 C) 0.12 D) 0.045 E) 0.053
please help!!

Answer 1

2010 (9) +MxH4(9) = 3N2 (9) +2H2009) Initially 0.08 mol 0.35 mol 0 0 change zu u +34 +24 Equilibrium (0-08-24) (0:35-x) +3% +2x At equilibrium number of moles of N, O = 0.050 mol So, 0-08-2x = 0.050 21 = 0.08-0.050 2K = 0.03 n = 0.015 mol +3k +2K Number of moles of Ny at equilibrium = 37 = 3 (0.015) mol 0.045 mol option D.