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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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Answer:

Mass percent Nā‚‚ = 89%

Mass percent Hā‚‚ = 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 atmĀ·LĀ·molā»Ā¹Ā·Kā»Ā¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolHā‚‚ + MolNā‚‚ = 0.307 mol

and

  • MolHā‚‚ * 2 g/mol + MolNā‚‚ * 28 g/mol = 3.49 g

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

Express MolHā‚‚ in terms of MolNā‚‚:

  • MolHā‚‚ + MolNā‚‚ = 0.307 mol
  • MolHā‚‚ = 0.307 - MolNā‚‚

Replace that value in the second equation:

  • MolHā‚‚ * 2 g/mol + MolNā‚‚ * 28 g/mol = 3.49
  • (0.307-MolNā‚‚) * 2 + MolNā‚‚ * 28 = 3.49
  • 0.614 - 2MolNā‚‚ + 28molNā‚‚ = 3.49
  • 0.614 + 26MolNā‚‚ = 3.49
  • MolNā‚‚ = 0.111 mol

Now we calculate MolHā‚‚:

  • MolHā‚‚ + MolNā‚‚ = 0.307 mol
  • MolHā‚‚ + 0.111 = 0.307
  • MolHā‚‚ = 0.196 mol

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

  • Nā‚‚ ā‡’ 0.111 mol * 28 g/mol = 3.108 g Nā‚‚
  • Hā‚‚ ā‡’ 0.196 mol * 2 g/mol = 0.392 g Hā‚‚

Mass % Nā‚‚ = 3.108/3.49 * 100% = 89.05% ā‰… 89%

Mass % Hā‚‚ = 0.392/3.49 * 100% = 11.15% ā‰… 11%

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