dr 4 1IS required to rully react with all of your CaCl2. 6. (10 pts) In your lab
ID: 545451 • Letter: D
Question
dr 4 1IS required to rully react with all of your CaCl2. 6. (10 pts) In your laboratory notebook write a procedure for how you will isolate and weigh the precipitate. 7. (4 pts) Record the masses of the precipitate formed during each reaction. 8. (4 pts) Record all observations you make during the experiment. 9. (10 pts) Organize your data into a table that includes all the class data WEEK 2 10. (20 pts) Organize the class data so that is visually represents the fixed mass ratio of calcium chloride to sodium phosphate. (HINT: Read the background carefully) 11. (10 pts) Write a conclusion in your notebook. Your conclusion should be based on the data from the entire class. In your conclusion you must argue either for or against the law of definite proportions and provide the mass ratio. Laboratory Experiment 5
Explanation / Answer
10)First write a balanced equation for the reaction:
3 CaCl2 + 2 Na3PO4 ----> Ca3 (PO4)2 + 6 NaCl
3moles CaCl2 will react with 2 moles of Na3PO4
Molar mass CaCl2 = 110.98 g/mol
3moles of CaCl2 = 3 x 110.98g/mol = 332.94g
Molar mass Na3PO4 = 163.94 g/mol
2 moles of Na3PO4 = 2 x 163.94 = 327.88g
332.94g CaCl2 will react with 327.88g Na3PO4
1g CaCl2 will react with 327.88/ 332.94 g Na3PO4
1g CaCl2 will react with 0.985g Na3PO4
Molar mass of Ca3 (PO4)2 =310.18 g/mol
1 molCa3 (PO4)2 = 310.18 g
327.88g /310.18 g/mol Ca3 (PO4)2 = 0.946 g Ca3 (PO4)2
1g CaCl2 will react with 0.985g Na3PO4, produces 0.946 g Ca3 (PO4)2
Group
CaCl2(g)
Na3PO4(g)
Ca3 (PO4)2(g)
1
0.15
0.05
0.048
2
0.15
0.10
0.096
3
0.15
0.15
0.144
4
0.15
0.20
0.143
5
0.15
0.25
0.143
For Group1
g of precipitate (Ca3 (PO4)2)(0.05g x (0.946 g/0.985g)) = 0.048g
For Group2
g of precipitate (Ca3 (PO4)2) = (0.10 g x 0.946g/0.985g) = 0.095
For Group3
g of precipitate (Ca3 (PO4)2) = (0.15 x (0.946g/0.985g) ) =0.144
For Group4
g of precipitate (Ca3 (PO4)2) = (0.15g x 0. 0.946g/1g) =0.143
For Group5
g of precipitate (Ca3 (PO4)2) = (0.15g x 0. 0.946g/1g) = 0.143
11) The law of definite proportions states that a given chemical compound always contains the same elements in the exact same proportions by mass, so it is against the law of definite proportions’
Law of multiple proportions states when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.
For example, there are three distinct weight of Na3PO4 0.05g, 0.1 g and 0.15g in combination with 0.15 grams of CaCl2 gives, Ca3 (PO4)2 in increasing order 0.048,0.096, 0.144grams, or in a ratio of 1, 2, 3.
Group
CaCl2(g)
Na3PO4(g)
Ca3 (PO4)2(g)
1
0.15
0.05
0.048
2
0.15
0.10
0.096
3
0.15
0.15
0.144
4
0.15
0.20
0.143
5
0.15
0.25
0.143
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