2NOBr <--> 2NO + Br2?

a sample of NOBr (.64mol) was placed in a 1 L flask containing no NO or Br2. At equilibrium the flask contained .46 mol of NOBr. How many moles of NO and Br2, respectively, are in the flask at equilibrium? no clue how to do this problem. thanks

Answer:
initial concentration NOBr = 0.64 / 1 = 0.64 M
concentration NOBr at equilibrium = 0.46 / 1 = 0.46 M
2 NOBr <-----> 2 NO + Br2
initial concentration
0.64 . . . . .. . . . .0. . .. . .0
change
.-x . . . . . . . . . . +x . . . . .+x/2
x = 0.64 - 0.46 = 0.18
at equilibrium
0.46 . . . . . . . . . 0.18 . . . 0.09
Concentration NO = 0.18 M ; moles = molarity x Volume = 0.20 moles / L x 1L = 0.18 moles
Similary moles Br2 = 0.09
The balanced formula is your clue. 2 molecules of NOBr yield 2 molecules of NO and 1 molecule of Br2.

0.64-0.46 moles ( 0.18 moles) of NOBr disappeared so you expect to get 0.18 moles of NO and 0.09 moles of Br2 at equilibrium
this question has very little to do with equilibrium as it has to do with stoichiometry. lets up a chart. what we know ill put in brackets and then explain later.

2NOBr --> 2NO + Br2

[0.64mol]...0...0

-0.18..+.18...+0.09

[.46]...0.18.0.09

the NOBr started out as 0.64 moles and became 0.46 moles. that means thant it went down 0.18 moles. since the NO also has a 2 in front of it, it goes up the same amout as the NOBr went down. reactants are used up and products are formed. the Br2 goes up 1/2 of what the NOBr goes down since it has a 1 in front of the compound.
Your biggest clue is the amount of NOBr at equilibrium which is given already at 0.46 moles. This means that the original 0.64 moles is decreased by this amount. 0.64- 0.46 = 0.18 moles.

Then, look at the mole ratios of the balanced equation. This means 2 moles of NOBr will yield 2 moles of NO and 1 mole of the Br2. Correspondingly, by ratio and proportion, you get 0.18 moles of NO and .18/2 or 0.09 moles of Br2
at equilibrium.

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