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25 December, 19:32

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.

N2H4 (aq) + O2 (g) - --> N2 (g) + 2H2O (l)

If 2.05g of N2H4 reacts and produces 0.550 L of N2, at 295K and 1.00 atm, what is the percent yield of the reaction?

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Answers (2)
  1. 25 December, 23:04
    0
    Answer: 35.5%

    Explanation:

    First we need to find the number of mole of 2.05g of N2H4

    Molar Mass of N2H4 = (14x2) + (1x4)

    = 28 + 4 = 32g/mol

    Mass conc. of N2H4 = 2.05g

    Number of mole = Mass conc / Molar Mass

    Number of mole = 2.05/32 = 0.0641mol

    Next, We need to find the volume (theoretical yield) occupied by this mole (0.0641mol) of N2H4, using the ideal gas equation.

    n = 0.0641mol

    P = 1atm

    T = 295K

    R = 0.082atm. L/mol / K

    V = ?

    PV = nRT

    1 x V = 0.0641 x 0.082 x 295

    V = 1.5506L

    The volume (1.5506L) obtained is the theoretical yield

    But the experimental volume = 0.550 L

    Percentage yield = (Experimental yield / theoretical yield) x100

    = (0.550/1.5506) x 100

    = 35.5%
  2. 25 December, 23:04
    0
    The percent yield of the reaction is 35 %

    Explanation:

    In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.

    Let's verify the moles that were used in the reaction.

    2.05 g. 1mol / 32 g = 0.0640 mol

    In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.

    Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).

    1atm. 0.550L = n. 0.082. 295K

    (1atm. 0.550L) / 0.082. 295K = n → 0.0225 moles

    Percent yield of reaction = (Real yield / Theoretical yield). 100

    (0.0225 / 0.0640). 100 = 35%
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