Question

-5.95g of cobalt(II) carbonate were added to 40cm3 of hydrochloric acid, concentration 2.0mol/dm3
.
I] Calculate the maximum yield of cobalt(II) chloride-6-water and show that the cobalt(II) carbonate was in excess.

CoCO3 + 2HCl → CoCl2 + CO2 + H2O
CoCl2 + 6H2O → CoCl2.6H2O

} maximum yield:
} number of moles of HCl used =
} number of moles of CoCl2 formed =
} number of moles of CoCl2.6H2O formed =

} mass of one mole of CoCl2.6H2O = 238g
}maximum yield of CoCl2.6H2O = …………………………………………………..…..…g
to show that cobalt(II) carbonate is in excess:
number of moles of HCl used = ……………………….……… (use your value from above)

mass of one mole of CoCO3 = 119g

number of moles of CoCO3 in 5.95g of cobalt(II) carbonate = …………………………

(ii) Explain how these calculations show that cobalt(II) carbonate is in excess.

Answer 1

Answer on the Question #67257, Chemistry / Inorganic chemistry

5.95g of cobalt(II) carbonate were added to 40cm³ of hydrochloric acid, concentration 2.0mol/dm³

[I] Calculate the maximum yield of cobalt(II) chloride-6-water and show that the cobalt(II) carbonate was in excess.

$$\mathrm{CoCO_3} + 2\mathrm{HCl} \rightarrow \mathrm{CoCl_2} + \mathrm{CO_2} + \mathrm{H_2O}$$ $$\mathrm{CoCl_2} + 6\mathrm{H_2O} \rightarrow \mathrm{CoCl_2 \cdot 6H_2O}$$

maximum yield:

number of moles of HCl used =

number of moles of CoCl₂ formed =

number of moles of CoCl₂·6H₂O formed =

mass of one mole of CoCl₂·6H₂O = 238g

maximum yield of CoCl₂·6H₂O = ...g

to show that cobalt(II) carbonate is in excess:

number of moles of HCl used = ... (use your value from above)

mass of one mole of CoCO₃ = 119g

number of moles of CoCO₃ in 5.95g of cobalt(II) carbonate = ...

(ii) Explain how these calculations show that cobalt(II) carbonate is in excess.

Solution:

(i) Maximum yield of CoCl₂·6H₂O:

Number of moles of HCl used:

$$n(HCl) = \frac{c(HCl) \cdot V(HCl)}{1000} = \frac{2.0 \cdot 40}{1000} = 0.08 \text{ mole}$$

Number of moles of CoCl₂ formed:

$$n(CoCl_2) = \frac{1}{2} n(HCl) = 0.04 \text{ mole}$$

Number of moles of CoCl₂·6H₂O formed:

$$n(CoCl_2) = n(CoCl_2 \cdot 6H_2O) = 0.04 \text{ mole}$$

Maximum yield of CoCl₂·6H₂O:

$$m(CoCl_2 \cdot 6H_2O) = n \cdot M = 0.04 \cdot 238 = 9.52 \text{ g}$$

To show that Cobalt (II) carbonate is in excess:

Number of moles of HCl used = 0.04 mole

Number of moles of CoCO₃ in 5.95g of cobalt(II) carbonate:

$$n(CoCO_3) = \frac{m(CoCO_3)}{M(CoCO_3)} = \frac{5.95}{119} = 0.05 \text{ mole}$$

(ii) Cobalt (II) carbonate is in excess because of mole number of this compound is more than HCl.

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