Question

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

Answer 1

1533.6 kg NO

Step-by-step 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 *
`(4kmolNH_3)/(5kmolO_2)`) 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 *
  `(4kmolNO)/(4kmolNH_3)` = 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