contestada

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

Respuesta :

maacastrobr

Answer:

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 * [tex]\frac{4molNH_3}{5molO_2}[/tex] = 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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