Question

2 NH3 + 5 F2 → N₂F4 + 6 HF

If some amount of NH3 and F2 were allowed to react and 92 g of N₂F4 was formed, and
54 g of F₂ was still present,
What was the limiting reactant?
What mass of NH3 was originally present?
What mass of F2 was originally present?
What mass of HF was formed?

Answer 1

Limiting reagent is NH3
Mass of NH3 = 30.08 gm
Mass of F2 = 222.08 gm
Mass of HF = 106.15 gm

Step-by-step explanation:
Given reaction: 2 NH3 + 5 F2 → N₂F4 + 6 HF
2 moles of NH3 (17 u) react with 5 moles of F2 (38 u)
Now, we know that 54 gm of F2 was left over, hence the limiting reagent must be NH3.
So we shall use gravimetric analysis on NH3.

Molar mass of N2F4 = 104 u
Weight of N2F4 = 92 g
Moles of N2F4 = 92/104 moles
2 moles NH3 gives 1 mole N2F4
so 92/104 mole of N2F4 is given by 92*2/104 mole NH3.
184/104 mole NH3, or 184*17/104 = 30.08 g
The moles of F2 will be 92*5/104, and mass will be
168.08 + 54 = 222.08 gm

Mass of HF present will be 92*6*20/104 = 106.15 gm