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27 January, 13:48

Consider the following balanced equation:

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

If 70.8 moles of CuO (s) and 40.1 moles of NH3 (g) are allowed to react, what is the theoretical yield of Cu (s) in moles?

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  1. 27 January, 16:55
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    The theoretical yield of Cu (s) in moles is 60.15 moles

    Explanation:

    Step 1: Data given

    Number of moles CuO = 70.8 moles

    Number of moles NH3 = 40.1 moles

    Molar mass CuO = 79.545 g/mol

    Molar mass NH3 = 17.03 g/mol

    Step 2: The balanced equation

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

    For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

    NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles

    There will remain 70.8 - 60.15 = 10.65 moles CuO

    Step 3: Calculate moles Cu

    For 3 moles CuO we need 2 moles NH3 to produce 3 moles H2O, 3 moles Cu and 1 mol N2

    For 40.1 moles NH3 we'll have 60.15 moles Cu

    The theoretical yield of Cu (s) in moles is 60.15 moles
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