Question

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

Answer 1

Answer:

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%