I think he did explain it well in the video, so you can watch that again if you want. So you have the Xe and F2 which forms XeF2. You're told that 30cm^3 of xenon and 20cm^3 of fluorine react to form X amount of XeF2. The equation is already balanced, so we know that 1 mole of xenon and 1 mole of F2 will form 1 mole of XeF2. Since the xenon is in excess, all of the fluorine and xenon have to react to form the 1 mole of XeF2, which has to be 20cm^3 because that is the lowest volume out of the 2 reactants, so the product formed would have to be 20cm^3. Then you would have 10cm^3 left of xenon because all of the fluorine reacted with all of the xenon to form the product, so the final answer is the volume of xenon + XeF2 which is 20 + 10 = 30cm^3.
in 17:05 would the oxidation state of nitrogen not be +5 instead of +3?
it is +5
how did you figure out for question 22 that 20cm3 to xef2 are produced?
I think he did explain it well in the video, so you can watch that again if you want. So you have the Xe and F2 which forms XeF2. You're told that 30cm^3 of xenon and 20cm^3 of fluorine react to form X amount of XeF2. The equation is already balanced, so we know that 1 mole of xenon and 1 mole of F2 will form 1 mole of XeF2. Since the xenon is in excess, all of the fluorine and xenon have to react to form the 1 mole of XeF2, which has to be 20cm^3 because that is the lowest volume out of the 2 reactants, so the product formed would have to be 20cm^3. Then you would have 10cm^3 left of xenon because all of the fluorine reacted with all of the xenon to form the product, so the final answer is the volume of xenon + XeF2 which is 20 + 10 = 30cm^3.
thank u so much sir
True saviour. I got my exams in a week and these are really helpful
Thanks :-). Good luck with your exams.
Very very useful thing to do, I've just discovered your channel and I'm so grateful
Thanks for that nice comment. Males it worthwhile making them :-)
amazing, could u do paper 2 2016 aswell please
Thanks for your kind comment! I've just uploaded paper 2 now.