Question

When ammonia (NH3) reacts with fluorine, the products are gaseous dinitrogen tetrafluoride (N2F4) and hydrogen fluoride (HF).

b) How many moles of each reactant are needed to produce 2.96 moles of HF?

c)How many grams of F2 are required to react with 40.2 g of NH3?

d)How many grams of N2F4 can be produced when 5.71 g of NH3 reacts?

Answer 1

Balanced equation :

2NH3 + 5F2 -> N2F4 + 6HF

b) According to the reaction,

6 mole HF require 2 mole NH3 and 5 mole F2

2.96 moles of HF require 2×2.96/6 = 0.987 moles NH3

and 5×2.96/6 = 2.47 moles of F2

c) According to the reaction,

2mole NH3 require 5 mole F2

2×17 = 34 g NH3 require 5×38 = 190 g F2

40.2 g NH3 require 190×40.2/34 g F2 = 224.65 g F2

d) According to the reaction,

2 mole NH3 give 1 mole N2F4

2×17 = 34 g NH3 give 104 g N2F4

5.71 g NH3 give 104×5.71/34 g N2F4

= 17.46 g N2F4