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14 November, 18:34

If 25.0 g of NH3 and 38.9g of O2 react in the following reaction, how

many grams of NO will be formed?

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H20 (g)

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Answers (1)
  1. 14 November, 20:25
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    There will be formed 36.49 grams of NO

    Explanation:

    Step 1: The balanced equation

    4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H20 (g)

    Step 2: Data given

    Mass of NH3 = 25.0 grams

    Mass of O2 = 38.9 grams

    Molar mass of NH3 = 17.03 g/mol

    Molar mass of 02 = 32 g/mol

    Molar mass of NO = 30.01 g/mol

    Step 3: Calculate number of moles

    Moles of NH3 = 25.0 grams / 17.03 g/mol

    Moles of NH3 = 1.47 moles

    Moles of O2 = 38.9 grams / 32 g/mol

    Moles of O2 = 1.216 moles

    Step 4: Calculate limiting reactant

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

    O2 is the limiting reactant. It will complete be consumed (1.216 moles).

    NH3 is the limiting reactant. There will 4/5 * 1.216 = 0.9728 moles of NH3 consumed.

    There will remain 0.4972 moles of NH3.

    Step 5: Calculates moles of NO

    If there is consumed 4 moles of NH3, there is produced 4 moles of NO

    For 1.216 moles of NH3 consumed, we will have 1.216 moles of NO

    Step 6: Calculate mass of NO

    Mass NO = moles NO * Molar mass NO

    Mass NO = 1.216 moles * 30.01 g/mol = 36.49 grams

    There will be formed 36.49 grams of NO
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