Asked by RM
6.0g of cobolt (II) carbonate was added to 40cm3 of hydrochloric acid, concentration 2.0 mol/dm3. Calculate the maximum yield of cobalt (II) choloride-6-water and show that cobalt (II) carbonate was in excess.
CoCO3 +2HCl = CoCl2 +CO2 + H2O
CoCl2 + 6H2O = CoCl2.6H2O
what is the number of HCL moles used?
CoCO3 +2HCl = CoCl2 +CO2 + H2O
CoCl2 + 6H2O = CoCl2.6H2O
what is the number of HCL moles used?
Answers
Answered by
MathMate
1. Calculate moles of CoCO3
=6.0/(58.933+12.0+3*16.0)
=6.0/118.9
=0.0504 mol
2. Calculate moles of HCl
=40/1000*2.0 mol
=0.080
Using stoiciometric ratios
CoCO3 used = 0.08*1/2=0.04 mols
=> CoCO3 is in excess (<0.0504).
The remaining answer can be obtained by finding the mass of 0.040 mol of CoCl2 and H2O respectively.
[assuming choloride-6-water meant
choloride-&-water]
=6.0/(58.933+12.0+3*16.0)
=6.0/118.9
=0.0504 mol
2. Calculate moles of HCl
=40/1000*2.0 mol
=0.080
Using stoiciometric ratios
CoCO3 used = 0.08*1/2=0.04 mols
=> CoCO3 is in excess (<0.0504).
The remaining answer can be obtained by finding the mass of 0.040 mol of CoCl2 and H2O respectively.
[assuming choloride-6-water meant
choloride-&-water]
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.