Wondering if someone could help me by looking over my work. I got a little confu

Question

Wondering if someone could help me by looking over my work. I got a little confused at the bottom when creating the pie charts for trial 2.

I.      PURPOSE (10 POINTS)

To conduct two double displacement reactions, I will determine the amount of calcium present when calcium chloride reacts with potassium carbonate. The calcium carbonate should precipitate out of the solution and the calcium carbonate will be determined through gravimetric analysis.

II.    TEST DATA (15 POINTS)

Trial 1

Mass of weigh boat: 1.6g

Mass of same weigh boat and CaCl2 (Note: This sample will be set aside and exposed to air for 24 hours): 3.6g

Mass of same weigh boat and CaCl2 after 24 hours: 5.0g

Observations of CaCl2: Small white spheres are solid at room temp. Highly soluble when added to distilled water.

Mass of 50 mL beaker: 7.2g

Mass of same beaker and K2CO3: 7.2g (beaker)+2.5g (K2CO3) = 9.7g

Amount of time beaker solution stirred: 4 minutes

Amount of time beaker solution set: 15 minutes

Mass of filter paper and watch glass: 1.1g(filter paper) + 33.6g(watch glass)=34.7g

Mass of same filter paper, watch glass, and dried product: 35.8g

Trial 2

Mass of 250 mL beaker: 117.8g

Mass of same beaker and CaCl2: 121.2 g

Observations of CaCl2: The calcium chloride had almost completely turned to a liquid state. There are only 5 solid pieces left.

Mass of 50 mL beaker: 7.2g

Mass of same beaker and K2CO3: 7.2g+2.5g(K2CO3)=9.7g

Amount of time beaker solution stirred: 4 minutes

Amount of time beaker solution set: 15 minutes

Mass of filter paper and watch glass: 1.1g(filter paper) + 33.6g(watch glass)=34.7g

Mass of same filter paper, watch glass, and dried product: 36.1g

III. CALCULATIONS (15 POINTS)

Trial 1

Mass of CaCl2: 2.0 g

Mass of K2CO3: 2.5 g

Mass of product: 1.1g

Mass of water absorbed with the CaCl2 after 24 hours: Mass of weigh boat with CaCI2 before absorption is 3.6g after 24hrs with water is 5g.

3.6g-5.0g = 1.4g water absorbed

Trial 2

Mass of CaCl2: 3.3g

Mass of K2CO3: 2.5g

Mass of product: 1.4g

IV. RESULTS (20 POINTS)

Trial 1

K2CO3 + CaCl2 -----> 2KCl + CaCO3

Theoretical yield (CaCO3):

1 mole of CaCO3 is formed from 1 mole of K2CO3

Mass of K2CO3 = 2.5 g

So moles of K2CO3 = 2.5 / MW of K2CO3 =2.5 / 138.205 = 0.018

So 0.018 moles of CaCO3 will be produced. This corresponds to moles x MW of CaCO3 = 0.018 x 100.0869 g

= 1.802 g

Actual yield (CaCO3):

1.1 g    

Percent yield:

So, % yield of CaCO3 = (actual yield / theoretical yield) x 100 % = (1.1 / 1.802) x 100 % = 61.04 %

Moles of Ca+2 present in original solution:

Moles of CaCl2 = mass of CaCl2 / MW of CaCl2 = 2.0 / 110.98 = 0.01802

As 1 mole of CaCl2 gives 1 mole of Ca2+, moles of Ca2+ = 0.01802 moles

Mass of CACl2 present in original solution:

2.0 g

Percent water absorbed by CaCl2:

Mass of water absorbed after 24 hr = 5.0

Explanation / Answer

In the trial 2 you have:

Trial 2

Mass of 250 mL beaker: 117.8g and

Mass of same beaker and CaCl2: 121.2 g

the difference is 121.2 g - 117.8 g = 3.4 g  but you said you have 3.3 g. Check this.

the amount of water absorbed by CaCl2 shouldn`t affect the outcome if you take the weight of CaCl2 fast enough before it start to take much water. Remember that you filter the CaCO3 and if you dried well this don't sum to your solid. So you can't justify don't reach a yield of 100 % because of the water taken by CaCl2, (At least it already had water when you weighed it). If CaCl2 already had water when you weighed then your explanation is right.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.