4NH3 + 502 --> 4NO + 6H20 How much excess reactant is leftover after if 6.30g of ammonia react with
1.80g of oxygen?

5.52 g

Explanation:

4NH₃ + 5O₂ → 4NO + 6H₂O

First we convert the given masses of both reactants into moles, using their respective molar masses:

6.30 g NH₃ ÷ 17 g/mol = 0.370 mol NH₃ 1.80 g O₂ ÷ 32 g/mol = 0.056 mol O₂

Now we calculate with how many NH₃ moles would 0.056 O₂ moles react, using the stoichiometric coefficients:

0.056 mol O₂ * = 0.045 mol NH₃

As there more NH₃ moles than required, NH₃ is the excess reactant.

Then we calculate how many NH₃ moles remained without reacting:

0.370 mol NH₃ - 0.045 mol NH₃ = 0.325 mol NH₃

Finally we convert NH₃ moles into grams:

0.325 mol NH₃ * 17 g/mol = 5.52 g

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