In lab we mixed a solution of powdered Calcium Chloride Dihydrate with DI water

Question

In lab we mixed a solution of powdered Calcium Chloride Dihydrate with DI water and .500 M sodium carbonate together, heated it, chilled it, filtered it, and dried the precipitate formed.

My data consists of
Mass of Calcium Chloride Dihydrate Mixture: 10.582 g- 7.710g= 2.872g
Mass of precipitate formed: 2.241g- 1.017g= 1.224g

1) Use the mass of calcium carbonate and the molar mass of calcium carbonate to calculate the number of moles of calcium carbonate:
CaCo3 molar mass: 100.09g
1.224g x (1mol/100.09g) = 0.01222899 mol CaCO3

2) Write out the chemical equation for the precipitate reaction used in the experiment and balance it.
CaCl2*2H2O(s) + Na2CO3 (aq) -> CaCO3 (s) + 2NaCl (s) + 2H2O (l)

Questions Needing Help:

3) Find the mol of calcium chloride dihydrate present in the sample of the mixture
- is it asking from the precipitate?
- Since the molar ration is 1:1 does that mean the number of moles is the same as the calcium carbonate?

4) Calculate the mass of CaCl2*2H2O present in the sample of the mixture used in each trial

5) Using the mass of CaCl2*2H2O and the mass of the mixture, calculate the mass % of calcium chloride dihydrate in the mixture)

7) Find the mass % of NaCl in the mixture using the mass % of calcium chloride dihydrate .

Explanation / Answer

3) Looking at the reaction we can say that, 1 mole of CaCO3 is formed from 1 mole of CaCl2.2H2O

So, moles of CaCl2.2H2O = moles of CaCO3 = 1.22 x 10^-2 mols

4) mass of CaCl.2H2O = moles x molar mass = 1.22 x 10^-2 x 147.0146 = 1.7936 g

5) mass % of CaCl2.2H2O in the mixture = (1.7936/2.241) x 100 = 80.03%

6) moles of NaCl = 2 x moles of CaCO3 = 2 x 1.22 x 10^-2 = 0.0244 mols

mass of NaCl = 0.0244 x 58.44 = 1.426 g

mass % of NaCl in the mixture = (1.426/2.241) x 100 = 63.63%

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