Question

Post Lab Questions:

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

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

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

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

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

Based on these findings:

Data Table 1. Stoichiometry Values

Initial: CaCl2•2H2O (g)	.68
Initial: CaCl2•2H2O (moles)	.0068
Initial: CaCl2 (moles)	.00368
Initial: Na2CO3 (moles)	.0012
Initial: Na2CO3 (g)	.12
Theoretical: CaCO3 (g)	.68
Mass of Filter paper (g)	1g
Mass of Filter Paper + CaCO3 (g)	1.7g
Actual: CaCO3 (g)	.72g
% Yield:	103

Answer 1

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

Solution :-

If the reaction is not gone for the completion then it will not form the complete precipitate and some of the reactant may lost in the solution while the filtration this will give the lower percent yield.

If the precipitate is not dried well and contains the moisture then it will give the higher mass of the precipitate because of the added moisture this will give the higher percent yield than actual.

If the precipitate is lost during the filtration process then it will also gives the lower percent yield.

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

Solution :- CaCl2 is the limiting reactant so adding more Na2CO3 will not have any affect on the formation of the precipitate so the percent yield will not affected by the amount of the Na2CO3 if it is added in more amount.

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

Solution :-

Mass of CaCl₂•9H₂O = 273.1215 g/mol

Mass of CaCl2 = 110.9840 g/mol

(7.5 g CaCl₂•9H₂O * 110.9840 g CaCl2 / 273.1215 g CaCl₂•9H₂O) = 3.05 g CaCl2

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

Solution :-

Molar mass of MgSO₄ = 120.3676 g/mol

Molar mass of MgSO₄•7H₂O = 246.4746 g/mol

(2.4 g MgSO₄•7H₂O * 120.3676 g MgSO4 / 246.4746 g MgSO₄•7H₂O) = 1.17 g MgSO4

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

Solution :- To find the conservation of the mass in the experiment we can calculate the initial mass of the CaCl2 that is used and from that we can find the mass of Ca initially used and then find the mass of the final product and from the mole ratio we can find the mass of Ca in the product and compare it with the initial mass of the Ca which was used in the starting of the experiment.