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Answered

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

AN2
BNBr3
CNaOH
DHOBr

Verified Answer :

Answer:

The correct answer is option B.

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:

\(\frac{2}{3}\times 48 mol=32 mol\) of \(NBr_3\)

Then ,40 moles of \(NaBr_3\) will reacts with:

\(\frac{3}{2}\times 40 mol=60 mol\) 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.

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 \(\frac{48}{3}\times 2=32\) 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) \(NBr_3\)

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