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20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of NO. A. yes B. false

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  1. Yesterday, 14:44
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    There will be produced 30.64 grams of nitrogen dioxide

    The statement is false

    Explanation:

    Step 1: Data given

    MAss of nitrogen produced = 20.0 grams

    Molar mass of N2 = 28.0 g/mol

    Mass of NO = 20.0 grams

    Molar mass of NO = 30.01 g/mol

    Volume of O2 at STP = 29.8 L

    Step 2: the balanced equation

    2NO + O2 → 2N02

    Step 3: Calculate moles NO

    Moles NO = mass NO / molar mas NO

    Moles NO = 20.0 grams / 30.01 g/mol

    Moles NO = 0.666 moles

    Step 4: Calculate moles O2

    1 mol O2 at STP = 22.4 L

    29.8 L = 29.8/22.4 = 1.33 moles

    Step 5: Calculate the limiting reactant

    For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

    NO is the limiting reactant. IT will completely be consumed (0.666 moles).

    O2 is in excess. There will react 0.666/2 = 0.333 moles O2

    There will remain 1.333 - 0.333 = 0.997 moles O2

    Step 6: Calculate moles NO2

    For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

    For 0.666 moles NO we'll have 0.666 moles NO2

    Step 7: Calculate mass NO2

    Mass NO2 = moles NO2 * molar mass NO2

    Mass NO2 = 0.666 moles * 46.0 g/mol

    Mass NO2 = 30.64 grams

    There will be produced 30.64 grams of nitrogen dioxide

    The statement is false
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