Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT. 10.00 g of O2 reacts with 20.00 g NO

Question

Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.

10.00 g of O2 reacts with 20.00 g NO

The molar masses of O2, NO, and NO2 are 32.00 g/mol, 30.01 g/mol, and 46.01 g/mol, respectively

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Thanh Hà 3 years 2021-09-02T23:27:22+00:00 1 Answers 4 views 0

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  1. mit
    0
    2021-09-02T23:28:31+00:00
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    Answer:

    balanced equation mole ratio 5 2 mol NO/1 mol O2

    10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

    20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

    actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

    Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

    ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

    Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

    0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

    Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

    Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

    Explanation:

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