#1
Okay i have a pretty long Q here and i'm stuck on a few bits of it, if you can read through and help i'll be majorly greatful (the bits i'm stuck on are highlighted in red)

Ammonium Sulphate reacts with aqueous sodium hydroxide as shown in the EQ
-
(NH4)2SO4 + 2NaOH -> 2NH3 + Na2SO4 + 2H2O

A sample of Ammonium sulphate was heated with 100 cm of 0.500moll/dm3 aquous Sodium Hydroxide (this was in excess)
Heating was continued until all the ammonia had been driven off as a gas. The unreacted NaOH required 27.3cm3 of 0.600 mol/dm3 HCL for neutralisation

a) Calc. the original number of moles of NaOH in 100 cm of 0.500 mol/dm3 NaOH
-Mol (NaOH) = 0.5 x 0.1 = 0.05 moles

b)Calc no. of mol of HCL in 27.3 of 0.600mol/dm3 HCL
-Mol (HCL)= 0.6 x 0.0273 = 0.0164 moles

c)Deduce the no. of moles of the unreacted NaOH neutralised by the HCL?

Okay i have this as something like HCL divided by NaOH so : 0.0164/0.05= 0.328 moles however this is more than i started with and leads me to believe i'm mega wrong

d) Use you're answers from A and C to calc. the number of moles of NaOH which react with ammonium sulphate?

Errr....
#2
for c you need to know whether the reaction is one to one, one to two etc, and the apply the ratio to the moles of HCL. For one to one for example it would be 0.0164 moles. for d you do original moles A, minus unreacted moles C to get reacted moles. Simple as
#5
no, for C you need to use the reaction between NaOH and HCL -----> NaCl + H20
This is one to one, so the moles NaOH neutralised in this will be equal to the moles HCL, which as you said is 0.0164
#7
i was just going to say that lol.

and for d you just add them together.


learn this format, theres a few questions i've seen exactly like this
Get off this damn forum and play your damn guitar.