Random Problems 21. In an experiment, 100.0 mL of silver nitrate of unknown conc
ID: 568392 • Letter: R
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Random Problems 21. In an experiment, 100.0 mL of silver nitrate of unknown concentration and 100.0 mL of calcium chloride of unknown concentration are mixed together. A white precipitate forms. When 10.0 mL more calcium chloride is added, no more precipitate forms. The solution is then filtered and the precipitate is weighed to be 2.15 g. Then excess 0.500 M sodium carbonate is added to the previous solution. Another precipitate forms. When it is filtered, the new precipitate weighs 2.33 g. What are the concentrations of the silver nitrate and calcium chloride solutionsExplanation / Answer
In the formation of first precipitate, the limiting reagent is AgNO3 (since CaCl2 is excess)
moles of AgCl formed = weight/ MW = 2.15/143.32 = 0.015
2 AgNO3(aq)+ CaCl2(aq) ------->2 AgCl (s) + Ca(NO3)2(aq)
2 mole of AgCl produced from 2 mole of AgNO3 and 1 moles of CaCl2
0.015 mole of AgCl produced from ---?mole of AgNO3 and ---? moles of CaCl2
Number of moles of AgNO3 = 0.015
Concentration of AgNO3 = moles/vol in lit = 0.015/0.1 = 0.0015 M = 1.5 x 10-3M
Number of moles of CaCl2 used in the precipitation of AgNO3 = 0.015 / 2 = 0.0075
In the formation of second precipitate, the limiting reagent is CaCl2 (since Na2CO3 is excess)
Na2CO3(aq)+ CaCl2(aq) -------> 2 NaCl aq) + CaCO3(s)
weight of CaCO3 =2.33 grams
Number of moles of CaCO3= weight/MW = 2.33/100 =0.0233
1 mole of CaCO3 produced from 1 mole of Na2CO3 and 1 moles of CaCl2
0.0233 mole of CaCO3 produced from --? of CaCl2 = 0.0233
total moles of CaCl2 = 0.0233+0.0075 = 0.0308
Total volume of CaCl2 = 100 ml +10 ml = 110 ml =0.11 L
Concentration of CaCl2 = moles/vol in lit = 0.0308/0.11 = 0.0028 M = 2.8 x 10-3M
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