In preparation for a combustion demonstration, a professor 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?

For a combustion decomposition , 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 night , both the gases leak through pores in the balloon .

The % of H2 leaks = 35%

Moles of H2 leaks = 0.35 mol

moles of H2 remaining = 0.65 mol

The molecular mass of O2 = 32g

The molecular mass of H2 =2g

Thus, the rate of diffusion of O2 = R

and the rate of diffusion of H2 = r

Now , Graham's law of diffusion ,

[tex]R/r = \sqrt{\frac{M( hydrogen ) }{M( oxygen)} } =\sqrt{\frac{2}{32} } =\frac{1}{4}[/tex]

thus the diffusion rate of O2/H2 is 1/4 .

From this we get to know that the % of O2 leaks is 1/4 of H2 leaks .

% of O2 leaks = 1/4 *35% = 8.8%

Moles of O2 leaks = 0.088 mol

Thus , the remaining O2 in the balloon = 0.912 mol

Then , the ratio of O2/H2 in the balloon the next day = 0.912/0.65 =1.4

Hence , the ratio of O2/H2 the next day is 1.4 .

Learn more about Graham's law of diffusion here :

brainly.com/question/12415336

#SPJ4