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4 November, 18:51

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

N2H4 (aq) + O2 (g) ⟶N2 (g) + 2H2O (l) If 3.15 g of N2H4 reacts with excess oxygen and produces 0.950 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

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  1. 4 November, 19:27
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    Percent yield = 39.3%

    Explanation:

    Given dа ta:

    Mass of N₂H₄ = 3.15 g

    Volume of nitrogen produced = 0.950 L

    Temperature = 295 K

    Pressure = 1 atm

    Percent yield = ?

    Solution:

    Chemical equation:

    N₂H₄ + O₂ → N₂ + 2H₂O

    Number of moles of N₂H₄:

    Number of moles = mass / molar mass

    Number of moles = 3.15 g/32 g/mol

    Number of moles = 0.1 mol

    Now we will compare the moles of hydrazine and nitrogen.

    N₂H₄; N₂

    1 : 1

    0.1 : 0.1

    PV = nRT

    V = nRT/P

    V = 0.1 mol * 0.0821 atm. L/mol. K * 295 K / 1 atm

    V = 2.42 L

    Percent yield:

    Percent yield = actual yield / theoretical yield * 100

    Percent yield = 0.950 L / 2.42 L * 100

    Percent yield = 39.3%
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