|
|
Post by shanice on Nov 27, 2015 23:50:33 GMT
December 2014 Question no. 2  |
|
|
|
Post by Admin on Nov 28, 2015 7:16:50 GMT
Hi Shanice, i see the photo.
what is your question?
|
|
|
|
Post by shanice on Nov 28, 2015 15:44:48 GMT
I attempted the question and I'm not sure if I'm  |
|
|
|
Post by kiyajoestar on Nov 30, 2015 20:03:39 GMT
a) To find the amount of moles of excess H2S04 remaining after step 2, I found:
-the amount of moles in the titre value, 13.35ml of NaOH
-Noted that NaOH reacts with H2SO4 in a 2:1 mole ratio respectively.
-Found how much moles of H2SO4 would have reacted with the amount of NaOH in the titre value and got the amount of H2S04 Excess.
-I got 0.001335 moles excess H2SO4 , or in scientific notation, 1.3 x 10^-3 moles.
b)To find the # of moles of H2S04 consumed, I found:
- the # of moles in 50 ml
- subtracted the amount excess H2S04 from the amount in 50 ml
- I got 0.003665 moles H2S04 consumed, or 3.445 x 10^-3 moles.
c) To find the mass of CH3COONa present in the 10g of the powder, I found:
-The molar mass of CH3COONa= 82g
-Wrote an equation to determine the ratio CH3COONa reacts with H2S04 and got 2:1 mole ration respectively
-Used the # of moles H2S04 consumed and the mole ratio to determine the # of moles of CH3COONa the H2SO4 reacted with.
-noted that that is the # of moles in 25ml, and found for 250ml
-Found the mass of CH3COONa that would be equal to the number of moles in 250ml
-I got 6.012g CH3COONa in 10g of powder.
Not sure if it's correct but I spoke to someone else that got these answers as well, hope this helps =3
|
|
|
|
Post by shanice on Nov 30, 2015 22:23:17 GMT
Thanks a lot  |
|
megan
New Member
Posts: 1 
|
Post by megan on Dec 1, 2015 17:13:02 GMT
Sir, can you confirm this ? ^  |
|
|
|
Post by Admin on Dec 1, 2015 19:59:23 GMT
Hi , i did not calculate the values myself, but it looks like kiyajoestar's method is correct.
|
|
|
|
Post by brandon17ruben on Dec 1, 2015 20:42:21 GMT
For part (c), the equation would be H2SO4 + CH3COONa --------> CH3COOH + Na2SO4 + H2O, wouldn't the mole ratio be 1:1 instead of 2:1?
|
|
|
|
Post by Admin on Dec 1, 2015 21:17:31 GMT
For part (c), the equation would be H2SO4 + CH3COONa --------> CH3COOH + Na2SO4 + H2O, wouldn't the mole ratio be 1:1 instead of 2:1? Hi, water isn't formed in that reaction. the equation is H2SO4 + 2 CH3COONa --------> 2 CH3COOH + Na2SO4 |
|
