Question

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?

Answer 1

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]