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9 March, 08:25

A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the mass percent of these two gases. Assume ideal-gas behavior. slader

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  1. 9 March, 08:50
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    Mass percent N₂ = 89%

    Mass percent H₂ = 11%

    Explanation:

    First we use PV=nRT to calculate n, which is the total number of moles of nitrogen and hydrogen:

    1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K n = 0.307 mol

    So now we know that

    MolH₂ + MolN₂ = 0.307 mol

    and

    MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

    So we have a system of two equations and two unknowns. We use algebra to solve it:

    Express MolH₂ in terms of MolN₂:

    MolH₂ + MolN₂ = 0.307 mol MolH₂ = 0.307 - MolN₂

    Replace that value in the second equation:

    MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49 0.614 - 2MolN₂ + 28molN₂ = 3.49 0.614 + 26MolN₂ = 3.49 MolN₂ = 0.111 mol

    Now we calculate MolH₂:

    MolH₂ + MolN₂ = 0.307 mol MolH₂ + 0.111 = 0.307 MolH₂ = 0.196 mol

    Finally, we convert each of those mol numbers to mass, to calculate the mass percent:

    N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂ H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

    Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

    Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%
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