Given this data from a paper filter precipitation experiment, can you help me, p

Question

Given this data from a paper filter precipitation experiment, can you help me, please?

Initial: CaCl2•2H2O (g)

1.0

Initial: CaCl2•2H2O (moles)

0.068

Initial: CaCl2 (moles)

0.068

Initial: Na2CO3 (moles)

0.068

Initial: Na2CO3 (g)

.7209

Theoretical: CaCO3 (g)

0.680

Mass of Filter paper (g)

1

Mass of Filter Paper + CaCO3 (g)

1.7

Actual: CaCO3 (g)

.72

% Yield:

103%

a. A perfect percent yield would be 100%. Based on your results, describe your degree of accuracy and suggest possible sources of error.

B.      What impact would adding twice as much Na2CO3 than required for stoichiometric quantities have on the quantity of product produced?

C.       Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O.

D.      Determine the quantity (g) of pure MgSO4 in 2.4 g of MgSO4•7H2O.

E.       Conservation of mass was discussed in the background. Describe how conservation of mass (actual, not theoretical) could be checked in the experiment performed.

Initial: CaCl2•2H2O (g)

1.0

Initial: CaCl2•2H2O (moles)

0.068

Initial: CaCl2 (moles)

0.068

Initial: Na2CO3 (moles)

0.068

Initial: Na2CO3 (g)

.7209

Theoretical: CaCO3 (g)

0.680

Mass of Filter paper (g)

1

Mass of Filter Paper + CaCO3 (g)

1.7

Actual: CaCO3 (g)

.72

% Yield:

103%

Explanation / Answer

(A)

A possible source of error is the error which could have occurred during measurement of mass of filter paper. The apparatus might not have been calibrated properly, or some human error while reading the measurement.

(B)

No impact, because the product formed would be the same and would depend on the amount of limiting reagent and not on the one which is present in excess.

(C)

Moles of CaCl2•9H2O = 7.5/273.12 = 0.0274

So, mass of pure CaCl2 = 0.0274*111 = 3.04 g

(D)

Moles of MgSO4•7H2O = 2.4/246.5 = 0.0097

So, mass of pure MgSO4 = 0.0097*120.36 = 1.167 g

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