What is the mole fraction of O2 in a mixture of 15.1 g of O2, 8.18 g of N2, and 2.46 g of H2?

Sagot :

Solution:

Step 1: Calculate the molar mass of each gas.

For O₂

molar mass = 16.00 g/mol × 2

molar mass = 32.00 g/mol

For N₂

molar mass = 14.01 g/mol × 2

molar mass = 28.02 g/mol

For H₂

molar mass = 1.008 g/mol × 2

molar mass = 2.016 g/mol

Step 2: Calculate the number of moles of each gas.

[tex]n_{\text{O}_{2}} = \text{15.1 g O₂} × \frac{\text{1 mol O₂}}{\text{32.00 g O₂}} = \text{0.472 mol}[/tex]

[tex]n_{\text{N}_{2}} = \text{8.18 g N₂} × \frac{\text{1 mol N₂}}{\text{28.02 g N₂}} = \text{0.292 mol}[/tex]

[tex]n_{\text{H}_{2}} = \text{2.46 g H₂} × \frac{\text{1 mol H₂}}{\text{2.016 g H₂}} = \text{1.22 mol}[/tex]

Step 3: Calculate the mole fraction of O₂.

[tex]X_{\text{O}_{2}} = \frac{n_{\text{O}_{2}}}{n_{\text{O}_{2}} + \: n_{\text{N}_{2}} + \: n_{\text{H}_{2}}}[/tex]

[tex]X_{\text{O}_{2}} = \frac{\text{0.472 mol}}{\text{0.472 mol + 0.292 mol + 1.22 mol}}[/tex]

[tex]\boxed{X_{\text{O}_{2}} = 0.238}[/tex]

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