Question

If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

A) N2
B) NBr3
C) NaOH
D) HOBr

Answer 1

Answer : The correct option is (B)
`NBr_3`

Solution : Given,

Moles of
`NBr_3` = 40 mol

Moles of
`NaOH` = 48 mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2NBr_3+3NaOH\rightarrow N_2+3NaBr+3HOBr`

From the balanced reaction we conclude that

As, 3 mole of
`NaOH` react with 2 mole of
`NBr_3`

So, 48 moles of
`NaOH` react with
`(48)/(3)* 2=32` moles of
`NBr_3`

From this we conclude that,
`NBr_3` is an excess reagent because the given moles are greater than the required moles and
`NaOH` is a limiting reagent and it limits the formation of product.

Excess moles of
`NBr_3` = 40 - 32 = 8 moles

Hence, the correct option is (B)
`NBr_3`

Answer 2

The correct answer is option B.

Step-by-step explanation:

`3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2`

Moles of
`NBr_3` = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles
`NBr_3`

Then ,48 moles of NaOH will reacts with:

`(2)/(3)* 48 mol=32 mol` of
`NBr_3`

Then ,40 moles of
`NaBr_3` will reacts with:

`(3)/(2)* 40 mol=60 mol` of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the
`NBr_3` is an excessive reagent.