The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.
1 answer:
5F2 + 2NH3 --> N2F4 + 6HF
<span>60.1g NH3 / 17g/mole = 3.54moles NH3 </span>
<span>3.54moles NH3 x (5 F2 / 2NH3) x 38g/mole = 335.85g required </span>
<span>5.25g HF / 20g/mole = 0.262moles HF </span>
<span>0.262moles HF x (2NH3 / 6HF) x 17g/mole = 1.49g required </span>
<span>209g / 38g/mole = 5.5moles F2 </span>
<span>5.5moles F2 (1 N2F4 / 5F2) x 66g/mole = 72.6g produced </span>
<span>Li3N + 3H2O --> NH3 + 3LiOH </span>
<span>(37.7g / 34.7g/mole) x (3H2O / 1 Li3N) x 18g/mole = 58.67g required </span>
<span>1.08moles Li3N (1NH3 / 1Li3N) x 6.022x10^23molecules/mole = 6.54x10^23 molecules </span>
<span>10.3L at STP: 10.3L / 22.4L/mole = 0.46moles NH3 produced </span>
<span>0.46moles NH3 x (1Li3N / 1NH3) x 34.7g/mole = 15.96g</span>
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