Recent content by jackthehat

I noted that the number of carbons in the main chain is 7 (5 carcons and the outer two are attached to an outside 'CH3' grouping .. and the bonds are all single so we have a Heptane. We have the main chain diagram structure as follows .. -C-C-C-C-C-C-C- .. If we then number these carbons...

Hi DrClaude, Thank you for all the help you gave me .. they say that you only really learn by your mistakes and I think that I now have a better understanding of things. Better late than never I suppose. Thanks agin for all your help. Regards, Jackthehat

Hi again, I applied the 90% immediately after I calculated the number of moles for each of the reactants .. the actual figures for each substance before the reaction .. the figures for NH3 and O2 before I applied the 90% were .. NH3 = 48.42 moles O2 = 90.79 moles When do I apply the 90% then...

Hi again DrClaude, I am sorry for the confusion I have caused on this .. the amounts before the reaction and the amounts after the 90% yield reaction that I estimated are as follows ... Substance Before reaction After 90% yieldd C3H6 30.255...

Hi DrClaude, Thanks for getting back to me .. I will list all of the amounts for each of the reactants and products below .. The mole amounts I got (for a 90% yield reaction) are as follows ... C3H6 = 16.335 moles O2 = 40.082 moles C3H6 = 3.026 moles C3H3N = 27.234...

Hi again everyone, I went over my problem again taking into account all of what I hope I have learned in this forum from all of you who have helped and tackled the problem again from the begining. I calculated the number of mole of C3H6 (Limiting Reagent) that theoretically would react...

Hi mjc123. Thank you for that explanation .. I see where I am going wrong I'm think now. As I said earlier although I had been marked wrong for my solution I was given no feedback as to why and I wanted to understand the problem and solution more clearly. Just one little thing I would like to...

Hi there, Thank you for replying to my query .. Now you say that my above method of solving the problem is completely wrong .. and you seem to infer that my understanding of 'yield' in a chemical reaction is not correct .. do you mean that 90% yield is not a reduced amount of product after a...

Hi Borek, Thank you once again for taking the time to help me. A couple of things .. Sorry should have explained more clearly why I calculated mass of product earlier. There were two parts to this question .. first part specifically asked for the mass of product synthesised by this reaction for...

I had already found the Mass of the product (C3H3N) produced by this reaction (theoretical mass at 100% yield) in a previous problem. I did this by finding the Limiting Reagent (C3H6) in the reaction , calculating the number of moles of C3H6 and using the Molar Ratios in the balanced reaction...

Hi Borek, Thank you for taking the time and effort to answer my problem. I was really confused as to what ratio I had to use to find the percentage yield, and I think that was because I hadn't really understood what percentage yield really meant. That it was, as you state, " the real over...

Hi BvU, Thank you for taking the time to look at my problem and give some input. Do you mean that the limiting reagent produces the maximum mass of product we are able to produce in this reaction and so we have to calculate how much excess reagent mass it would take to produce 163g of product...

Summary:: We are given a description of a reaction and the masses of two reactants and are initially asked to calculate which of two is the excess reagent and how much of the excess reagent would be left over if we have 100% completion. Then we are given a final mass (after purification) of the...

Hi mjc123, Thanks a lot for taking the time to help me with my problem. Regards, Jack

Hi Borek, Thank you very much for your help .. I have to say at one point i never thought I'd get there but I suppose the old adage is apt .." If at first you don't succeed, try, try and try again." Regards, Jack