Answer:
The correct answer is option B.
Explanation:
[tex]3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2[/tex]
Moles of [tex]NBr_3[/tex] = 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles [tex]NBr_3[/tex]
Then ,48 moles of NaOH will reacts with:
[tex]\frac{2}{3}\times 48 mol=32 mol[/tex] of [tex]NBr_3[/tex]
Then ,40 moles of [tex]NaBr_3[/tex] will reacts with:
[tex]\frac{3}{2}\times 40 mol=60 mol[/tex] 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 [tex]NBr_3[/tex] is an excessive reagent.
Answer : The correct option is (B) [tex]NBr_3[/tex]
Solution : Given,
Moles of [tex]NBr_3[/tex] = 40 mol
Moles of [tex]NaOH[/tex] = 48 mol
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2NBr_3+3NaOH\rightarrow N_2+3NaBr+3HOBr[/tex]
From the balanced reaction we conclude that
As, 3 mole of [tex]NaOH[/tex] react with 2 mole of [tex]NBr_3[/tex]
So, 48 moles of [tex]NaOH[/tex] react with [tex]\frac{48}{3}\times 2=32[/tex] moles of [tex]NBr_3[/tex]
From this we conclude that, [tex]NBr_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]NaOH[/tex] is a limiting reagent and it limits the formation of product.
Excess moles of [tex]NBr_3[/tex] = 40 - 32 = 8 moles
Hence, the correct option is (B) [tex]NBr_3[/tex]
