Answered

The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.

5 F2(

g. + 2 NH3(

g. N2F4(

g. + 6 HF(

g. (

AIf you have 73.7 g NH3, how many grams of F2 are required for complete reaction?

BHow many grams of NH3, are required to produce 5.25 g HF?

(

CHow many grams of N2F4 can be produced from 269 g F2?

Verified Answer :

5F2 + 2NH3 --> N2F4 + 6HF 
60.1g NH3 / 17g/mole = 3.54moles NH3 
3.54moles NH3 x (5 F2 / 2NH3) x 38g/mole = 335.85g required 

5.25g HF / 20g/mole = 0.262moles HF 
0.262moles HF x (2NH3 / 6HF) x 17g/mole = 1.49g required 

209g / 38g/mole = 5.5moles F2 
5.5moles F2 (1 N2F4 / 5F2) x 66g/mole = 72.6g produced 


Li3N + 3H2O --> NH3 + 3LiOH 
(37.7g / 34.7g/mole) x (3H2O / 1 Li3N) x 18g/mole = 58.67g required 

1.08moles Li3N (1NH3 / 1Li3N) x 6.022x10^23molecules/mole = 6.54x10^23 molecules 

10.3L at STP: 10.3L / 22.4L/mole = 0.46moles NH3 produced 
0.46moles NH3 x (1Li3N / 1NH3) x 34.7g/mole = 15.96g

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