AQA CHEM1: 15th May 2012

Totally overthought 5 (c) part i :/ where it said work out the conc of HCL, i worked out the moles of hcl using the moles of the reactant like everyone else but for the volume of the HCL i read the question and it said something along the lines of "the reaction produces 500cm3 of solution" i took that to mean the other aqueous product as well

becuase there was one molecule of that other product and three of the HCL i divided the 500 by 4 and multiplied by three to give what i thought the volume of the HCL would be, but ive spoke to a few people and noone else has done that

dont suppose anyone else done this? :P

isn't it giant metallic

5a) Work out total moles of gas on right hand side using pV=nRT. It was (5000cm3) so convert to m3.
Then divide by 5 the get one mole of B2O3.
Mr (69.6) x Mole = 2.81g i believe?

you were up till 4? :O

me too 2.21 as was to 3sf

You just use the number of moles of the reactant u worked out, times it by 3 to give moles of HCL, then divide by (500/1000)

H2BO3 + NaOH -> NaHBO3 + H20

Yeah i did something similar, on the right hand side the ratio of moles was 1 : 3 (HCL). Therefore volume of HCL was 375cm3.
Work out moles on the left, multiply by 3 to get moles of HCL. c=n/v convert the 375cm3 to dm3 by dividing by 1000. answer i got was 2.95g..

Yeah i did something similar, on the right hand side the ratio of moles was 1 : 3 (HCL). Therefore volume of HCL was 375cm3.
Work out moles on the left, multiply by 3 to get moles of HCL. c=n/v convert the 375cm3 to dm3 by dividing by 1000. answer i got was 2.95g..

how many people actually divided by 5 in that pv=nrt question? :L

Nope 14.1g.

i think we're right

i think people forgot to divide 500cm-3 by 1,000,000 to get M-3

That was my answer! glad i wasnt the only one!! looks like we might have got it wrong though :/

NO! what on earth is giant about it? got to the aqa textbook ?

I got somthing like 14!!!
Yess why the fk people are people getting '2.' somthing