Trial 1. Mass of CaCl2 2.0g Mass of K2CO3 2.5g Mass of filter paper 1.2g Mass of

Question

Trial 1. Mass of CaCl2 2.0g

              Mass of K2CO3 2.5g

              Mass of filter paper 1.2g

              Mass of Glass watch 37.5g

              Mass of Product 1.6g

Trial 2. Mass of CaCl2 (after 24hr exposure to air) 3.7g

              Mass of K2CO3 2.5g

              Mass of filter paper 1.2g

              Mass of Glass watch 37.5g

              Mass of Product 1.5g

Trial 1:

1. Theoretical yield (CaCO3)? 1.8g

2. Actual yield (CaCO3)? 1.6g

3. Percent yield? 88.88%

4. Moles of Ca present in original solution, based on actual yield? 0.016 mol

5. Mass of CaCl2 present in original solution, based on actual yield? 1.77g

Trial 2:

1. Theoretical yield (CaCO3)? 1.8g

2. Actual yield (CaCO3)? 1.5g

3. Percent yield? 83.33%

4. Moles of Ca present in original solution, based on actual yield? 0.015 mol

5. Mass of CaCl2 present in original solution, based on actual yield? 3.08g

Lab Questions

1. Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in % wt/v concentration.

2. Determine the mass of water which became associated with the calcium chloride during its exposure to the air.

3. Create a bar graph comparing the supposed percentage concentration and the actual percentage concentration of calcium chloride in the solution.

4. Create a pie chart representing the percent by mass of calcium, chloride and water in the original solution.

Having an issue with this one, I have the gist of the graph portion in the lab questions but questions one and two within the lab portions are stumping me. Thanks in advance.

Explanation / Answer

lab question

1)trial 1

Mass of CaCl2 present in original solution, based on actual yield= 1.77g

moles CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles

Total Volume of solution =V (not mentioned in your data)

concentration of CaCl2 in wt/V%=1.77g/V (in ml)*100

2)trial 2

Mass of CaCl2 present in original solution, based on actual yield= 3.08 g

Mass of CaCl2 (after 24hr exposure to air) 3.7g

mass of water absorbed=3.7-3.08=0.62 g

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