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What mass of nitrogen monoxide is formed in the reaction of 869kg ammonia and 2480kg oxygen gas? (Please show work)

Sagot :

maacastrobr

Answer:

1533.6 kg NO

Explanation:

The reaction that takes place is:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we convert the masses of ammonia (NH₃) and oxygen gas (O₂) into moles, using their respective molar masses:

  • NH₃ ⇒ 869 kg ÷ 17 kg/kmol = 51.12 kmol NH₃
  • O₂ ⇒ 2480 kg ÷ 32 kg/kmol = 77.5 kmol O₂

77.5 kmol of O₂ would react completely with (77.5 kmol O₂ * [tex]\frac{4kmolNH_3}{5kmolO_2}[/tex]) 62 kmol of NH₃. There are not as many kmol of NH₃, so NH₃ is the limiting reactant.

Now we calculate how many kmol of NO are produced, using the limiting reactant moles:

  • 51.12 kmol NH₃ * [tex]\frac{4kmolNO}{4kmolNH_3}[/tex] = 51.12 kmol NO

Finally we convert kmol of NO to mass, using its molar mass:

  • 51.12 kmol NO * 30 kg/kmol = 1533.6 kg NO
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