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Damm [24]
2 years ago
15

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

Chemistry
1 answer:
USPshnik [31]2 years ago
6 0

Answer:

1533.6 kg NO

Explanation:

The reaction that takes place is:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

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

  • 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 * \frac{4kmolNH_3}{5kmolO_2}) 62 kmol of NH₃. There are not as many kmol of NH₃, so NH₃ is the limiting reactant.

Now we <u>calculate how many kmol of NO are produced</u>, using the <em>limiting reactant moles</em>:

  • 51.12 kmol NH₃ * \frac{4kmolNO}{4kmolNH_3} = 51.12 kmol NO

Finally we <u>convert kmol of NO to mass</u>, using its<em> molar mass</em>:

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

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Explanation:

Given the following data;

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To find the final pressure, we would use Gay Lussac's law;

Gay Lussac states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

PT = K

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Making P2 as the subject formula, we have;

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