Nitrosyl Bromide Decomposes according to:

2NOBr(g) <=> 2NO(g) + Br2(g)

*<=> is an equilibrium arrow(

1 mole each of NO and BR2 are in the flask. At equilibrium .60mol of BR2 remained.
What is the number of moles of NOBr and NO in the flask at equilibrium, respectively.

a) .2mol and .4mol
b).8mol and .6mol
c).8mol and .6mol
d)1.2mol and 1.2 mol
e).8mol and .2mol

.6 mole of Br? Then .4 moles of Br2 changed back, or .8 moles of NOBr now are in the flask, as is .2 moles of NO remaining.

Think that out.

To solve this problem, we need to use the relationship provided by the balanced equation for the decomposition of nitrosyl bromide:

2NOBr(g) <=> 2NO(g) + Br2(g)

We are given that initially, there is 1 mole each of NO and Br2 in the flask. At equilibrium, we are told that 0.6 moles of Br2 remain.

Since the ratio of moles between reactants and products in the balanced equation is 1:1, we can conclude that for every 1 mole of Br2 that decomposes, 1 mole of NOBr is consumed.

Therefore, the amount of NOBr that decomposes is equal to the amount of Br2 that remains, which is 0.6 moles.

Since 1 mole of NOBr decomposes to give 1 mole each of NO and Br2, we know that 0.6 moles of NOBr decomposes, resulting in the formation of 0.6 moles each of NO and Br2.

Finally, to find the moles of NO remaining at equilibrium, we subtract the moles of NO formed from the initial quantity of NO:

Initial moles of NO = 1 mole
Moles of NO formed = 0.6 moles
Moles of NO remaining at equilibrium = 1 mole - 0.6 moles = 0.4 moles

Therefore, at equilibrium, the number of moles of NOBr and NO in the flask are 0.8 moles and 0.4 moles, respectively.

The correct answer is a) 0.8 mol of NOBr and 0.4 mol of NO.