Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H2O that would form when 2.78 mol NH3 and 5.19 mol O2 react.

The amount of NO formed s 2.78 moles or 83.4 grams

The amount of H2O formed is 4.17 moles or 75.1 grams

Explanation:

Step 1: Data given

Ammonia = NH3

Oxygen = O2

nitrogen monoxide = NO

water = H2O

Number of moles NH3 = 2.78 moles

Number of O2 = 5.19 moles

Step 2: The balanced reaction

4NH3 + 5O2 → 4NO + 6H2O

Step 3: Calculate moles of products

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

NH3 is the limiting reactant

All the NH3 will react. There will be 0 moles of NH3 left

For 4 moles NH3 we need 5 moles O2

For 2.78 moles NH3 we need 5/4 * 2.78 = 3.475 moles

There will be left 5.19 - 3.475 = 1.715 moles O2

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 2.78 moles NH3 we'll have 2.78 moles NO and 6/4 * 2.78 = 4.17 moles H2O

Step 4: Calculate mass of NO and H2O

Mass = moles * molar mass

Mass NO = 2.78 moles * 30.01 g/mol

Mass NO = 83.43 grams

Mass H2O = 4.17 moles * 18.02 g/mol

Mass H2O = 75.14 grams

The amount of NO formed s 2.78 moles or 83.4 grams

The amount of H2O formed is 4.17 moles or 75.1 grams

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