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An important reaction in the production of nitrogen fertilizers is the oxidation of ammonia to form...

An important reaction in the production of nitrogen fertilizers is the oxidation of ammonia to form nitrogen monoxide and water. How many liters of nitrogen monoxide will be formed at the same time as 26,500L of water vapor if both are maintained at 500.0 degrees C and 762.5 mmHg?

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Answer #1

2NH3 (g) + 5/2O2 (g) ---------> 2NO (g) + 3H2O(g)

2 moles of ammonia produce two moles of nitrogen monoxide and three moles of water

1 mole of a gas = 22.4L

therefore when 3 X 22.4 L of water vapour are formed 2 X 22.4 L of NO are formed

lets equate the values

3 X 22.4 L of water = 2 X 22.4 L of NO

1L of H2O = 2 X 22.4 / 3 X 22.4 = 0.66 L of NO

It means when one 1.0L of H2O is formed 0.66L of NO are formed

When 26500L of water are formed, the NO formed = 26500 X 0.66 = 17490L

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