25. Given the reactions : Na2CO3(aq) + CaCl2 (aq) 2NaCl (aq) +CaCO3 (s) Na2CO3(a

Question

25. Given the reactions : Na2CO3(aq) + CaCl2 (aq) 2NaCl (aq) +CaCO3 (s)
Na2CO3(aq) + 2HCl CO2 + 2NaCl +H2O
If 250.0ml of 1.5 M Na2CO3 is added to 250.0ml of a CaCl2 solution with an unknown
concentration and the resulting precipitate is removed from the liquid. Assume the removal of
the precipitate does not result in a change in the volume of liquid. Given that an excess of
Na2CO3 was used, the resulting liquid is titrated with hydrochloric acid until the remaining
sodium carbonate was neutralized. If 47.35 ml of 0. 137 M HCl were required to neutralize the
excess Na2CO3 . Give your answers in the correct number of significant digits
a) Determine the molar concentration of Na2CO3 present in the solution after the
reaction with CaCl2
b) Determine the concentration of the CaCl2 solution used in the initial reaction
c) Determine the theoretical yield (mass) of the precipitate formed.

Explanation / Answer

Solution :-

Given the reactions :

Na2CO3(aq) + CaCl2 (aq) 2NaCl (aq) +CaCO3 (s)
Na2CO3(aq) + 2HCl CO2 + 2NaCl +H2O

250.0 ml 1.5 M Na2CO3

Lets calculate the initial moles of the Na2CO3

Moles = molarity * volume in liter

           = 1.5 mol per L * 0.250 L

          = 0.375 mol Na2SO4

Now lets calculate the moles of the HCl

Moles of HCl =0.137 mol per L* 0.04735 L = 0.00649 mol HCl

Now using the mole ratio lets calculate the moles of the Na2CO3 reacted with the moles of the HCl

0.00649 mol HCl * 1 mol Na2CO3/2 mol HCl = 0.003245 mol Na2CO3

Now lets calculate the moles of the Na2CO3 reacted with CaCl2

Moles of Na2CO3 reacted with CaCl2 = 0.375 mol – 0.003245 mol = 0.3718 mol Na2CO3

Mole ratio of the Na2CO3 and CaCl2 is 1 : 1

Therefore moles of CaCl2 reacted = 0.3718 mol CaCl2

a)Determine the molar concentration of Na2CO3 present in the solution after the
reaction with CaCl2

Solution :- Na2CO3 moles present after reaction = 0.003245 mol

So molarity of the Na2CO3 after reaction is as follows

Volume = 250 ml +250 ml = 500 ml = 0.500 L

[Na2CO3]= moles/ volume in liter

                 = 0.003245 mol / 0.500 L

                 = 0.00649 M

b) Determine the concentration of the CaCl2 solution used in the initial reaction
Solution :-

moles of CaCl2 reacted = 0.3718 mol

therefore concentration of the CaCl2 = 0.3718 mol / 0.250 L = 1.487 M

c)Determine the theoretical yield (mass) of the precipitate formed.

Solution :-

Moles of CaCO3 formed = 0.3718 mol

Mass of CaCO3 = moles * molar mass

                            =0.3718 mol * 100.0869 g per mol

                             = 37.2 g

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