Calculate the number of NaBr formula units formed when 50 NBr3 molecules and 57 NaOH formula units react?

The reaction between NBr3 and NaOH can be written as:

`2NBr_3 + 3NaOH -> 3NaBr + N_2 + 3HOBr`

In this reaction, 2 moles of NBr3 react with 3 moles of NaOH to produce 3 moles of NaBr.

Here, instead of moles, we are given the amounts of the reactants in terms of number of molecules and formula units. These can also be treated in the same way as moles, as long as we remember...

The reaction between NBr3 and NaOH can be written as:

`2NBr_3 + 3NaOH -> 3NaBr + N_2 + 3HOBr`

In this reaction, 2 moles of NBr3 react with 3 moles of NaOH to produce 3 moles of NaBr.

Here, instead of moles, we are given the amounts of the reactants in terms of number of molecules and formula units. These can also be treated in the same way as moles, as long as we remember that a mole of a substance contains 6.023 x 10^23 molecules. 

2 moles of NBr3 reacts with 3 moles of NaOH.

or, 1 mole NBr3 reacts with 3/2 = 1.5 moles of NaOH.

or, 50 molecules of NBr3 will react with 50 x 1.5 = 75 molecules of NaOH.

However, we only have 57 formula units of NaOH. Thus, sodium hydroxide (NaOH) is the limiting chemical and will determine how much product is formed.

Since 3 moles of NaOH results in 3 moles of NaBr

thus, 57 formula units of NaOH will result in 57 formula units of NaBr. 

Hope this helps.