Nitric monoxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO₂), a dark brown gas. If 5.895 mol of NO is mixed with 2.503 mol of

Question

Nitric monoxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO₂), a dark brown gas. If 5.895 mol of NO is mixed with 2.503 mol of O₂,

determine the limiting reagent.

calculate the number of grams of NO₂ produced.

and determine how many grams of excess reagent remain unreacted.

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Jezebel 7 months 2021-07-11T13:36:36+00:00 1 Answers 2 views 0

Answers ( )

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    2021-07-11T13:38:29+00:00

    Answer:

    Limiting reactant: O2

    grams NO2 produced = 230.276 g NO2

    grams of NO unused = 26.67 gNO

    Explanation:

    2NO + O2 –> 2NO2

    Step 1: Determine the molar ratio NO:O2

    molar ratio NO:O2 = 5.895: 2.503 = 2.35

    stoichiometric molar ratio NO:O2 = 2:1

    So, O2 is the limiting reactant.

    Step2: Determine the grams of NO2:

    ?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2

    Step 3: Determine the amount of excess reagent unreacted

    moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted

    moles NO unreacted = total moles NO – moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted

    mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )