Nitrosyl bromide decomposes according to the following equation.

2NOBr (g)(equilibrium arrow) 2NO (g) + Br2 (g)

A sample of NOBr (0.64 mol) was placed in a 1.00-L flask containing no NO or Br2. At equilibrium
the flask contained 0.46 mol of NOBr. How many moles of NO and Br2, respectively, are in the
flask at equilibrium?

A) 0.18, 0.18 B) 0.18, 0.090 C) 0.46, 0.46 D) 0.46, 0.23 E) 0.18, 0.360

i thought it was .18 .18 but i was wrong. how is this problem solved/

It's 0.18 and 0.09

..........2NOBr ==> 2NO + Br2
initial....0.64..... 0.....0
change.....-2x.......+2x...+x
equilibrium. 0.46

0.64-0.46 = 0.18 and that is the same as NO (because the coefficients are equal) but the Br2 is 1/2 that. Or to say it another way.
2x = 0.64-0.46 = 0.18
x = 0.09
then 2NO = 2x = 0.18
x = Br2 = 0.09

Ah, chemistry problems. They can be as tricky as figuring out the correct punchline for a joke. But fear not, dear question asker, for I am here to help!

To solve this problem, we can use the fact that the molar ratios of the reactants and products remain constant during a chemical reaction. In this case, the ratio of NOBr to NO to Br2 is 1:2:1.

At the start, we have 0.64 mol of NOBr and no NO or Br2. Since the ratio of NOBr to NO is 1:2, we can calculate the amount of NO by dividing 0.64 mol by 2, giving us 0.32 mol of NO.

Now let's find the amount of Br2. Since the ratio of NOBr to Br2 is 1:1, the amount of Br2 is the same as the amount of NOBr, which is 0.46 mol.

So, to answer your question, at equilibrium, there are 0.32 moles of NO and 0.46 moles of Br2 in the flask. Therefore, the correct answer is option D) 0.46, 0.23.

I hope that clears things up and brings a smile to your face. Remember, chemistry can be fun, just like a good joke!

To solve this problem, we can start by setting up a table to keep track of the moles of each substance involved.

Let's denote the initial moles of NOBr as [NOBr]_initial, the moles of NO as [NO]_equilibrium, and the moles of Br2 as [Br2]_equilibrium.

Using the information given, we have:
[NOBr]_initial = 0.64 mol
[NOBr]_equilibrium = 0.46 mol

Since 2 moles of NOBr decompose to form 2 moles of NO and 1 mole of Br2, we can write the following equation:

[NO]_equilibrium = 2 * [NOBr]_initial - 2 * [NOBr]_equilibrium
[NO]_equilibrium = 2 * 0.64 mol - 2 * 0.46 mol
[NO]_equilibrium = 1.28 mol - 0.92 mol
[NO]_equilibrium = 0.36 mol

[Br2]_equilibrium = [NOBr]_initial - [NOBr]_equilibrium
[Br2]_equilibrium = 0.64 mol - 0.46 mol
[Br2]_equilibrium = 0.18 mol

Therefore, the moles of NO and Br2 at equilibrium are 0.36 mol and 0.18 mol, respectively.

The correct answer is option E) 0.18, 0.360.

To solve this problem, we can use the stoichiometry of the balanced chemical equation.

First, we need to calculate the change in moles of NOBr in the reaction. We know that the initial moles of NOBr is 0.64 mol, and at equilibrium, it decreases to 0.46 mol. Therefore, the change in moles of NOBr is:

Change in moles of NOBr = Initial moles of NOBr - Equilibrium moles of NOBr
= 0.64 mol - 0.46 mol
= 0.18 mol

Now, according to the balanced chemical equation, the stoichiometric ratio between NOBr and NO is 2:2 or 1:1. This means that for every 1 mole of NOBr that decomposes, 1 mole of NO is produced.

Therefore, the moles of NO at equilibrium is also 0.18 mol.

Similarly, the stoichiometric ratio between NOBr and Br2 is 2:1. This means that for every 2 moles of NOBr that decompose, 1 mole of Br2 is produced.

Therefore, the moles of Br2 at equilibrium is half the moles of NOBr that decomposed, which is 0.18 mol / 2 = 0.09 mol.

So, the correct answer is B) 0.18 mol of NO and 0.090 mol of Br2.