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5 April, 20:23

Consider the balanced equation for the following reaction:

3CuO (s) + 2NH3 (g) → 3H2O (l) + 3Cu (s) + N2 (g)

If 17.3 grams of CuO reacts with 81.4 grams of NH3, how much excess reactant remains?

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  1. 6 April, 00:02
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    There will remain 78.9 grams of NH3.

    Explanation:

    Step 1: Data given

    Mass CuO = 17.3 grams

    Molar mass CuO = 79.545 g/mol

    Mass NH3 = 81.4 grams

    Molar mass NH3 = 17.03 g/mol

    Step 2: The balanced equation

    3CuO (s) + 2NH3 (g) → 3H2O (l) + 3Cu (s) + N2 (g)

    Step 3: Calculate moles CuO

    Moles CuO = mass CuO / molar mass CuO

    Moles CuO = 17.3 grams / 79.545 g/mol

    Moles CuO = 0.217 moles

    Step 4: Calculate moles NH3

    Moles NH3 = 81.4 grams / 17.03 g/mol

    Moles NH3 = 4.78 moles

    Step 5: Calculate the limiting reactant

    CuO is the limiting reactant. It will completely react. NH3 is in excess. There will react 2/3 * 0.217 = 0.145 moles. There will remain 4.78 - 0.145 = 4.635 moles NH3

    Step 6: Calculate mass NH3 remaining

    Mass NH3 = 4.635 moles * 17.03 g/mol

    Mass NH3 = 78.9 grams

    There will remain 78.9 grams of NH3.
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