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31 July, 21:49

1. If 3.00g of ammonia reacts with 40.00g of oxygen, what is the theoretical yield of nitrogen monoxide?

2. If only 10.00g of nitrogen monoxide was isolated from the products, what was the percent yield of nitrogen monoxide?

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  1. 31 July, 21:56
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    The theoretical yield of nitrogen monoxide = 5.28 grams

    Explanation:

    Step 1: Data given

    Mass of ammonia = 3.00 grams

    Molar mass NH3 = 17.03 g/mol

    mass of oxygen = 40.00 grams

    Molar mass O2 = 32.0 g/mol

    Step 2: The balanced equation

    4NH3 + 5O2 → 4NO + 6H2O

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles NH3 = 3.00 grams / 17.02 g/mol

    Moles NH3 = 0.176 moles

    Moles O2 = 40.00 grams / 32.0 g/mol

    Moles O2 = 1.25 moles

    Step 4: Calculate limiting reactant

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

    NH3 is the limiting reactant. It will completely be consumed (0.176 moles)

    O2 is in excess. There will react 5/4 * 0.176 = 0.220 moles

    There will remain 1.25 - 0.22 = 1.03 moles

    Step 5: Calculate moles NO

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

    For 0.176 moles NH3 we'll have 0.176 moles NO

    Step 6: Calculate theoretical yield of NO

    Mass NO = moles NO * molar mass NO

    Mass NO = 0.176 moles * 30.01 g/mol

    Mass NO = 5.28 grams

    Step 7: Calculate percent yield

    % yield = (10.00 grams / 5.28 grams) * 100 %

    % yield = 189 %

    ⇒ Since the theoretical yield is less than 10.0 grams. I think the data are not correct (may be 30.00 grams of NH3)
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