In preparation for a combustion demonstration, a professor’s assistant fills a balloon with equal molar amounts of H2 and O2, but the demonstration has to be postponed until the next day. During the night, both gases leak through pores in the balloon. If 35% of the H2 leaks, what is the O2/H2 ratio in the balloon the next day? I calculated the diffusion ratio of O2/H2 and I get 1/4. The answer is 1.4 but I need help figuring out what to do next?

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## Answers (1)

Let's say that the balloon originally contained 1 mole of both H2 and O2. If 35% of the H2 leaked out, then the remaining amount of H2 would be 0.65 moles. If the O2 leaks out at 1/4 of the H2 rate, then

1/4 x 35% = 8.8% of the O2 leaked out. that would leave 0.912 moles of O2 in the balloon.

The new O2/H2 mole ratio is 0.912 / 0.65 = 1.4