Molarity of the sodium carbonate solution used (molL) e.l Molarty of the calcium

Question

Molarity of the sodium carbonate solution used (molL) e.l Molarty of the calcium chlonde soluticon used (molL)_0 (00 Trial Trial 1 Trial 2 Sodium Carbonate Solution Volune of so dium carbonate solution used (mL) Moles of sodium carbonate used (mol) 50.0 mEF Calcium Chloride Solution Volume ofthe calcium chlonde solution used (mL) Moles of calcium chloride used (mol) Vo.onl so-oml ·D M o.0 Solid Formed Mass of watch glass and liter Paper Mass of precipitate, watch glass and fiter paperS Mass of precipitate &) Actual moles of precipitate (mol) Theoretical moles of product formed (mol Percent Yield Pp(5S.0g 55.8338 030. 333 0.0036 ol 0.0033 el 2.x 1667 64% General Observations: Overall Results Average Percent Yield Standard Deviation Percent Yield RSD ofPercent Yield (%) Class Average Percent Yield This document was prepared by D. Finneran, J. Tiemey

Explanation / Answer

For the sodium carbonate solution

Trial 1

number of moles of solution = Volume of solution (in L) * molarity (M) = 50/1000 * 0.110 = 0.0055 moles

Trial 2

number of moles of solution = Volume of solution (in L) * molarity (M) = 50/1000 * 0.110 = 0.0055 moles

For the Calcium chloride solution

Trial 1

number of moles of solution = Volume of solution (in L) * molarity (M) = 50/1000 * 0.100 = 0.0050 moles

Trial 2

number of moles of solution = Volume of solution (in L) * molarity (M) = 50/1000 * 0.100 = 0.0050 moles

Remaining all data is correct, for class average you need to take the observations of every student and then take the average of the same

Q1) Balanced Reaction

Na2CO3(aq) + CaCl2(aq) --------- 2NaCl(aq) + CaCO3(s)

Q2) The full ionic equation will be

2Na+(aq) + CO3(2-)(aq) + Ca(2+)(aq) + 2Cl-(aq) --------- 2Na+(aq) + 2Cl-(aq) + CaCO3(s)

Q3) Net ionic equation will be

Ca(2+)(aq) + CO3(2-)(aq) ---------- CaCO3(s)

Q4) Take the average value of the class by finding the results of all students and compare it with your data

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