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7 May, 03:55

If 2.25 g of NH 3 reacts with 3.38 g of O2 and produces 0.450 L of N 2 at 295 K and 1.00 atm, which reactant is limiting?

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  1. 7 May, 04:20
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    NH3 is the limiting reactant. O2 is in excess and there will remain 0.224 grams O2

    Explanation:

    Step 1: Data given

    Mass of NH3 = 2.25 grams

    Mass of O2 = 3.38 grams

    Volume of N2 = 0.450 L

    Temperature = 295 K

    Pressure = 1.00 atm

    Molar mass NH3 = 17.03 g/mol

    Molar mass O2 = 32 g/mol

    Molar mass of N2 = 28 g/mol

    Molar mass of H2O = 18.02 g/mol

    Step 2: The balanced equation

    4NH3 + 3O2 → 2N2 + 6H2O

    Step 3: Calculate moles NH3

    Moles NH3 = mass NH3 / molar mass NH3

    Moles NH3 = 2.25 grams / 17.03 g/mol

    Moles NH3 = 0.132 moles

    Step 4: Calculate moles O2

    Moles O2 = mass O2 / molar mass O2

    Moles O2 = 3.38 grams / 32 g/mol

    Moles O2 = 0.106 moles

    Step 5: Calculate limiting reactant

    For 4 moles NH3 we need 3 moles O2 to produce 2 moles N2 and 6 moles H2O

    NH3 is the limiting reactant. It will completely be consumed (0.132 moles)

    O2 is in excess. There will react 3/4 * 0.132 = 0.099 moles O2

    There will remain 0.106 - 0.099 = 0.007 moles O2

    This is 0.007 moles * 32 g/mol = 0.224 grams

    NH3 is the limiting reactant. O2 is in excess and there will remain 0.224 grams O2
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