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24 June, 01:30

4.35 L of N2, at 25 ∘C and 1.16 atm, is mixed with 2.20 L of O2, at 25 ∘C and 0.277 atm, and the mixture allowed to react. N2 (g) + O2 (g) →2NO (g) How much NO, in grams, is produced?

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  1. 24 June, 01:35
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    Mass = 1.5 g

    Explanation:

    Given dа ta:

    Volume of nitrogen = 4.35 L

    Temperature = 25°C

    Pressure = 1.16 atm

    Volume of oxygen = 2.20 L

    Temperature = 25°C

    Pressure = 0.277 atm

    grams of NO = ?

    Solution:

    Moles of Nitrogen:

    PV = nRT

    n = PV/RT

    n = 1.16 atm*4.35 L / 0.0821 atm. L. K⁻¹. mol⁻¹ * 298 K

    n = 5.046 atm. L / 24.47 atm. L. mol⁻¹

    n = 0.206 mol

    Moles of oxygen:

    PV = nRT

    n = PV/RT

    n = 0.277 atm*2.20 L / 0.0821 atm. L. K⁻¹. mol⁻¹ * 298 K

    n = 0.609 atm. L / 24.47 atm. L. mol⁻¹

    n = 0.025 mol

    Balanced chemical equation:

    N₂ + O₂ → 2NO

    Now we will compare the moles of NO with moles of oxygen and nitrogen.

    N₂ : NO

    1 : 2

    0.206 : 2*0.206 = 0.412 mol

    O₂ : NO

    1 : 2

    0.025 : 2*0.025 = 0.05 mol

    The number of moles produced by oxygen are less so it will limiting reactant.

    Mass of NO:

    Mass = number of moles * molar mass

    Mass = 0.05 mol * 30 g/mol

    Mass = 1.5 g
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