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12 July, 02:03

It is desired to produce 4.74 grams of nitrogen dioxide by the following reaction. If the percent yield of nitrogen dioxide is 76.0 %, how many grams of nitrogen monoxide would need to be reacted? grams nitrogen monoxide nitrogen monoxide (g) + oxygen (g) nitrogen dioxide (g)

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  1. 12 July, 05:34
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    We need 4.08 grams of NO to be reacted

    Explanation:

    Step 1: The balanced equation

    2NO (g) + O2 (g) →2NO2 (g)

    Step 2: Data Given

    Mass of NO2 produced = 4.74 grams = actual yield

    percent yield = 76.0 %

    Molar mass of NO = 30.01 g/mol

    Molar mass O2 = 32 g/mol

    Molar mass NO2 = 46 g/mol

    Step 3: Calculate theoretical mass of NO2

    Yield = 0.76 = actual % / theoretical %

    Theoretical % = 4.74 / 0.76 = 6.24 grams

    Step 4: Calculate moles of NO2

    Moles of NO2 = mass NO2 / Molar mass NO2

    moles of NO2 = 6.24 grams / 46 g/Mol

    moles of NO2 = 0.136 moles

    Step 5: Calculate moles of NO

    For 2 moles of NO to be consumed, we have 2 moles of NO2 to be produced.

    So we have 0.136 moles of NO2 produced, and 0.136 moles of NO consumed

    Step 6: Calculate mass of NO

    mass = Moles NO * Molar mass NO

    mass NO = 0.136 mol * 30.01 g/mol = 4.08 grams

    We need 4.08 grams of NO to be reacted
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