• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Chemistry exam: Remember the question about a kid adding 1.35g (1 Viewer)

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
Remember the question that said a kid added 1.35g of something to 1.35g of something and he got copper or calcium carbonate, calculate how much of the copper/calcium carbonate was produced.

What were the actual elements anyone remeber their answers?
 

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
1.35g sodium carbonate and 1.35g copper nitrate were both placed into 2 separate beakers of water. These substances were then mixed together. What is the mass of the copper carbonate is formed? (not exact wording. I cut out 90% of it)

Do this please, show working.
 

Timothy.Siu

Prophet 9
Joined
Aug 6, 2008
Messages
3,449
Location
Sydney
Gender
Male
HSC
2009
1.35g sodium carbonate and 1.35g copper nitrate were both placed into 2 separate beakers of water. These substances were then mixed together. What is the mass of the copper carbonate is formed?

ok i'll try.
Na2CO3 + Cu(NO3)2 --->2NaNO3 + CuCO3

1.35/105.99=moles of sodium carbonate
1.35/187.57= moles of copper nitrate (limiting reagent?)
then u'll prolly get 1.35/187.57 moles of copper carbonate
wich is 1.35/187.57 x 123.56=0.889299994grams

i'm not sure...not too good with chem
 
Last edited:

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
Ok, now figure out the masss of copper carbonate.
 

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
I hope your right that what I put, but since I put it has to be wrong ( I mean me get something right, impossible):(
 

Timothy.Siu

Prophet 9
Joined
Aug 6, 2008
Messages
3,449
Location
Sydney
Gender
Male
HSC
2009
Shoom said:
I hope your right that what I put, but since I put it has to be wrong ( I mean me get something right, impossible):(
lol, then it mite be right coz i put it
 

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
If it is, I will be so happy I will smile. yes I rarely smile, I must be the most angry/dull person because I have a wrinkle on my forehead. Im noly 16 dammit, im gonna look like gordon ramsay when im 20!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 

Timothy.Siu

Prophet 9
Joined
Aug 6, 2008
Messages
3,449
Location
Sydney
Gender
Male
HSC
2009
Shoom said:
If it is, I will be so happy I will smile. yes I rarely smile, I must be the most angry/dull person because I have a wrinkle on my forehead. Im noly 16 dammit, im gonna look like gordon ramsay when im 20!!!!!!!!!!!!!!!!!!!!!!!!!!!!
it cudn't hav been worth that many marks, BE HAPPY!!! SMILE!!:D
 

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
I think it was worth 4 marks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:mad1: good this smiley has a wrinkle lol. :rofl:
 

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
Why did you use the Molar Mass of copper mitrate instead of the other thing? sorry for spelling im tired lol.
 

zzzz12345

Member
Joined
Jan 28, 2007
Messages
79
Gender
Female
HSC
2009
Shoom said:
1.35g sodium carbonate and 1.35g copper nitrate were both placed into 2 separate beakers of water. These substances were then mixed together. What is the mass of the copper carbonate is formed? (not exact wording. I cut out 90% of it)

Do this please, show working.
As stated before:
Na2CO3 + Cu(NO3)2 ---> CuCO3 + 2NaNO3

n(Na2CO3)= 1.35/105.99
= 0.0127 moles
n(Cu(NO3)2)=1.35/187.57
=0.007197
Therefore Cu(NO3)2 is the limiting reagent i.e. Copper nitrate will run out first in the reaction and thus no more product would be formed i.e. Copper Carbonate
Thus m(CuCO3)=0.007197 x (63.55+12.01+3 x 16)
=0.889 g (3dp)
You use the number of moles of copper nitrate because it is the limiting reagent (see explanation above). I suppose you can worry less now, if several people have gotten the same answer there is a high probability that you are right. Good Luck on your exams.
 

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
zzzz12345 said:
As stated before:
Na2CO3 + Cu(NO3)2 ---> CuCO3 + 2NaNO3

n(Na2CO3)= 1.35/105.99
= 0.0127 moles
n(Cu(NO3)2)=1.35/187.57
=0.007197
Therefore Cu(NO3)2 is the limiting reagent i.e. Copper nitrate will run out first in the reaction and thus no more product would be formed i.e. Copper Carbonate
Thus m(CuCO3)=0.007197 x (63.55+12.01+3 x 16)
=0.889 g (3dp)
You use the number of moles of copper nitrate because it is the limiting reagent (see explanation above). I suppose you can worry less now, if several people have gotten the same answer there is a high probability that you are right. Good Luck on your exams.
That's what I got.
 

Shoom

Member
Joined
Aug 26, 2008
Messages
694
Gender
Undisclosed
HSC
N/A
YYYYYYEEESSSSSSS!!!!!!!!!!!!!!!!!!!!

Thankyou all.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top