3N2H4 + 4ClF3 = 12HF + 3N2 +2Cl2
Intially: 0.880 atm N2H4, 0.970 atm CIF3.
At equilibrium Partial pressure N2 = 0.525 atm. Find partial pressure HF at equilibrium
...................3N2H4 + 4ClF3 = 12HF + 3N2 +2Cl2
I..................0.880.......0.970.........0.........0........0
C..................-3p............-4p........+12p...+3p....+2p
E...............0.880-3p..0.970-4p.....12p.....3p.....2p
The problem tells you that 3p = 0.525 so p must be 0.525/3 = ?