Question

4NH3 + 502 --> 4NO + 6H20

How much excess reactant is leftover after if 6.30g of ammonia react with
1.80g of oxygen?

Answer 1

5.52 g

Step-by-step 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 *
  `(4molNH_3)/(5molO_2)` = 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