gas laws

[jack_chay]

8.   During the electrolysis of dilute sulfuric acid solution, the gaseous electrode products are collected over water in separate gas burettes at 20 ° C and a total pressure of 101.3 kPa. At 20 ° C the vapour pressure of the water is 2.3 kPa. 0.20 g of hydrogen is collected.

(i)   What is the volume of hydrogen collected ?
(ii)   What is the mass of oxygen collected ?
   (iii) What is the volume of oxygen collected ?

ans

(i) 2.44 litres
      (ii) 1.59 grams
      (iii) 1.22 litres


thanks in advance

[BasicAcid]

Does pV=nRT work?

[jack_chay]

didn't know what to sub in as p for the first part?

[jack_chay]

that give wrong ans... i also triedd 99, that was wrong ans too

[--whiteskies]

that give wrong ans... i also triedd 99, that was wrong ans too

i tried it and got 2.404. i don't know - is that close enough? :'c

[psyxwar]

(i) 0.2 H₂ collected, n(H) = 0.2/2 = 0.1 mol


(ii) 4H⁺  + 2H₂O -> 2H₂ + O₂  + 4H⁺
Thus, for every one mole H₂ we produce half a mole of O₂.
n(O₂) = 0.05 mol
m(O₂) = 0.05 mol * 32gmol^-1 = 1.60g

(iii)

[SocialRhubarb]

Yeah, the answers I got are all slightly off as well - maybe their answers are based off a periodic table more accurate than ours, as the molar mass of hydrogen is actually slightly more than 1.

This is how I tried to solve it:
We have 0.20g of hydrogen gas, which is 0.10 moles of H₂. The partial pressure of this gas is 101.3kPa-2.3kPa=99kPa of the water, and we're assuming that no other gases are in the gas burette. This is the pressure we use and this gives us an answer of: 2.46L. Slightly off, as you've already said. However, if we use a slightly more accurate figure for hydrogen's molar mass of 1.008 amu, and hence a molar mass of 2.016 amu for the hydrogen molecule, we do get the given answer of 2.44L.

After this, we use the electrochemical series to look at the ratio of hydrogen gas to oxygen gas produced. The ratio of electrons accepted by the reduction of hydrogen gas and the electrons released by the oxidation of oxygen gas is 2:1, so the ratio of moles of hydrogen to oxygen gas is 2:1, giving us 0.05 moles of oxygen. I'm using the less accurate molar mass of hydrogen here. The mass of oxygen is then just the molar mass of the oxygen molecule multiplied by the moles of oxygen, which gives us 1.6g. This seems to be another error caused by the slightly different values for the mass of H.

Finally, the volume of oxygen produced, which we can calculate quite easily from the moles of oxygen we've already found, and given that you worked through the first part of the question just fine without me, I think you can handle it.

As a sidenote - you'll be using the value of 1.0 amu for molar mass of hydrogen tomorrow, so this sort of thing won't be an issue in the actual exam. Especially since you won't be able to check the answers.

[--whiteskies]

Yeah, the answers I got are all slightly off as well - maybe their answers are based off a periodic table more accurate than ours, as the molar mass of hydrogen is actually slightly more than 1.

This is how I tried to solve it:
We have 0.20g of hydrogen gas, which is 0.10 moles of H₂. The partial pressure of this gas is 101.3kPa-2.3kPa=99kPa of the water, and we're assuming that no other gases are in the gas burette. This is the pressure we use and this gives us an answer of: 2.46L. Slightly off, as you've already said. However, if we use a slightly more accurate figure for hydrogen's molar mass of 1.008 amu, and hence a molar mass of 2.016 amu for the hydrogen molecule, we do get the given answer of 2.44L.

oh i realised what i forgot to do, hence my 2.404. i forgot to subtract the 2.3kPa from 101.3kPa haha omg.