If the reaction was reversed and we wanted to produce as much NaN3, in grams, as possible from 30.0 g of N2 and 20.0 g of Na, which reactant

Question

If the reaction was reversed and we wanted to produce as much NaN3, in grams, as possible from 30.0 g of N2 and 20.0 g of Na, which reactant would be the limiting reactant? (4 points) How much NaN3 would actually be produced?

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Ngọc Diệp 4 years 2021-08-31T12:05:33+00:00 1 Answers 10 views 0

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  1. hongcuc2
    0
    2021-08-31T12:06:59+00:00
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    Answer:

    N₂

    Explanation:

    Sodium is a larger molecule with a much higher molecular weight. However, 20g of N₂ would be a smaller amount of molecules than 20g of sodium due to how there are multiple nitrogen molecules.

    2/3 * given mass of N₂ = mass of N₃

    N₃ (Azide ion) given mass = 20g

    Na = 20g

    Masses of chemicals are equal

    Na = 22.990g/mol

    20g/22.990g/mol = 0.8699mol of Na

    N₃ = 20g

    N₃ = N g/mol x 3

    N = 14.007 g/mol

    14.007 x 3 = 42.021 g/mol

    N₃ = 42.021g/mol

    20g/42.021g/mol = 0.4759 mol of N₃

    Notice how there are fewer moles of the Azide ion than the Sodium.

    0.4759 moles of NaN₃ is produced

    combine molecular weights:

    42.021 + 22.990 = 65.011 g/mol

    multiply by amount of moles of the limiting reactant:

    0.4759 mol *65.011 = 30.942 g

    Also, here is the balanced equation:

    3N₂ + 2Na = 2NaN₃

    The results are the same as the balanced equation.

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