Question

A mixture contains 20cm3 of hydrogen, 35cm3 of oxygen, 15cm3 of carbon monoxide and 10cm3 of nitrogen at STP. which of the following gives the mole fraction of hydrogen in this mixture?

Answer 1

Answer on Question #85196 – Chemistry – General Chemistry

A mixture contains $20~\mathrm{cm}^3$ of hydrogen, $35~\mathrm{cm}^3$ of oxygen, $15~\mathrm{cm}^3$ of carbon monoxide and $10~\mathrm{cm}^3$ of nitrogen at STP. Which of the following gives the mole fraction of hydrogen in this mixture?

Solution:

$\mathrm{V(H_2)} = 20~\mathrm{cm^3} = 20~\mathrm{mL} = 0.020~\mathrm{L}$

$\mathrm{n(H_2)} = \mathrm{V(H_2) / V_M(H_2)} = 0.020~\mathrm{L} / 22.4~\mathrm{L/mol} = 8.90 \times 10^{-4}~\mathrm{mol}$

$\mathrm{V(O_2)} = 35~\mathrm{cm^3} = 35~\mathrm{mL} = 0.035~\mathrm{L}$

$\mathrm{n(O_2)} = \mathrm{V(O_2) / V_M(O_2)} = 0.035~\mathrm{L} / 22.4~\mathrm{L/mol} = 1.56 \times 10^{-3}~\mathrm{mol}$

$\mathrm{V(CO)} = 15~\mathrm{cm^3} = 15~\mathrm{mL} = 0.015~\mathrm{L}$

$\mathrm{n(CO)} = \mathrm{V(CO) / V_M(CO)} = 0.015~\mathrm{L} / 22.4~\mathrm{L/mol} = 6.69 \times 10^{-4}~\mathrm{mol}$

$\mathrm{V(N_2)} = 10~\mathrm{cm^3} = 10~\mathrm{mL} = 0.010~\mathrm{L}$

$\mathrm{n(N_2)} = \mathrm{V(N_2) / V_M(N_2)} = 0.010~\mathrm{L} / 22.4~\mathrm{L/mol} = 4.46 \times 10^{-4}~\mathrm{mol}$

$\mathrm{n(mixture)} = 4.46 \times 10^{-4} + 6.69 \times 10^{-4} + 1.56 \times 10^{-3} + 8.90 \times 10^{-4} = 3.565 \times 10^{-3}~\mathrm{mol}$

$\chi(\mathrm{H_2}) = \mathrm{n(H_2)} / \mathrm{n(mixture)} = 8.90 \times 10^{-4} / 3.565 \times 10^{-3} = 0.25$

$\chi(\mathrm{H_2}) = 0.25$

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