Is the precision of your two trials within 5%. This means is your % average devi

Question

Is the precision of your two trials within 5%. This means is your % average deviation less than or equal to 5%? Calculations

Sample1Sample 2 1. Unknown CaCl, solution code (if any) Unknown Code wknown 2. Volume of 0.100 M Na PO, solution used . Initial burette volume of CaCl2 solution 4. Final burette volume of CaCla solution 5. Volume of unknown CaCl2 solution used 6. bo.usm 50.00 m 19.55m 29. 60 mL Per 49.1m1 Mass of filter paper ,485 7. Mass of filter paper + Cas(PO) precipitate 2.600 Mass of Cas(PO4)2 precipitate recovered o. b0l 9. Moles of Cas(POs)2 recovered 10. Moles of CaCl2 consumed in reaction 11. Molarity of unknown CaCl2 solution 0-003595m01 0.00194 mol 0.001m00 552 mo o 1o19 0.00 582 mo 0.29M 12. Average Value for Molarity of Unknown CaCl2 solution 0 415M Sample Calculations:

Explanation / Answer

The balanced chemical equation for the reaction between CaCl2 and Na3PO4 is given below.

3 CaCl2 (aq) + 2 Na3PO4 (aq) ----------> Ca3(PO4)2 (s) + 6 NaCl (aq)

As per the stoichiometric equation,

3 moles CaCl2 = 2 moles Na3PO4 = 1 mole Ca3(PO4)2

Trial 1

Trail 2

Mass of Ca3(PO4)2 precipitate recovered (g)

1.115

0.601

Mole(s) of Ca3(PO4)2 recovered = (mass of Ca3(PO4)2 recovered)/(molar mass of Ca3(PO4)2)

(1.115 g)/(310.18 g/mol) = 0.00359

(0.601 g)/(310.18 g/mol) = 0.00194

Mole(s) of CaCl2 consumed in reaction = (moles of Ca3(PO4)2)*(3 moles CaCl2/1 mole Ca3(PO4)2)

(0.00359 mole Ca3(PO4)2)*3 moles CaCl2/1 mole Ca3(PO4)2) = 0.01077

(0.00194 mole Ca3(PO4)2)*3 moles CaCl2/1 mole Ca3(PO4)2) = 0.00582

Trial 1

Trial 2

Mole(s) of CaCl2 consumed in the reaction

0.01077

0.00582

Mole(s) of Na3PO4 used in the reaction

(0.01079 mole CaCl2)*(2 moles Na3PO4/3 moles CaCl2) = 0.00719 mole.

(0.0582 mole CaCl2)*(2 moles Na3PO4/3 moles CaCl2) = 0.00388 mole.

Mole(s) of Na3PO4 taken initially = (volume of Na3PO4 taken in L)*(molarity of Na3PO4)

(30.0 mL)*(1 L/1000 mL)*(0.1 M) = 0.00300 mole

(30.0 mL)*(1 L/1000 mL)*(0.1 M) = 0.00300 mole

It seems that CaCl2 should be the excess reactant; the initial amount of Na3PO4 is less than the amount reacted and clearly, this means that Na3PO4 is the limiting reactant.

Mole(s) of Na3PO4 left unreacted = (moles of Na3PO4 taken initially) – (moles of Na3PO4 reacted)

N/A

N/A

Mass of unreacted Na3PO4 = (moles of unreacted Na3PO4)*(molar mass of Na3PO4)

N/A

N/A

The average molarity of the unknown CaCl2 is 0.415 M.

Standard deviation of the molarity of CaCl2 = [(0.5313 – 0.415)2 + (0.298 – 0.415)2]/(2 – 1) M = 0.1649 M.

The relative standard deviation of the molarity = (0.1649 M)/(0.415 M)*100% = 39.735%

The precision of the two trials is way more than 5% (ans).

Trial 1

Trail 2

Mass of Ca3(PO4)2 precipitate recovered (g)

1.115

0.601

Mole(s) of Ca3(PO4)2 recovered = (mass of Ca3(PO4)2 recovered)/(molar mass of Ca3(PO4)2)

(1.115 g)/(310.18 g/mol) = 0.00359

(0.601 g)/(310.18 g/mol) = 0.00194

Mole(s) of CaCl2 consumed in reaction = (moles of Ca3(PO4)2)*(3 moles CaCl2/1 mole Ca3(PO4)2)

(0.00359 mole Ca3(PO4)2)*3 moles CaCl2/1 mole Ca3(PO4)2) = 0.01077

(0.00194 mole Ca3(PO4)2)*3 moles CaCl2/1 mole Ca3(PO4)2) = 0.00582

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