3. This problem is similar to the analysis of the Za-Al alloy, excopt here you a

Question

3. This problem is similar to the analysis of the Za-Al alloy, excopt here you are dealing with reaction product that is a precipitate not a gas. A 5.000 gram sample of a mixture of NaCl and CaCl, was dissolved in pure water and the chloride ion was precipitated as silver chloride, AgCl. The dry AgCl precipitate weighed 12.554 8 Formula weights: AgCl= 143.4; NaCl= 58.5; Caci,-111.1 (a) Write the balanced chemical equations for the reactions of AgNO, with NaCI and with CaCl2 to produce the AgCl precipitate. (b) Write the algebraic equation that relates the number of mole of AgCl to the mole of NaCl and CaCl, in the original sample. NaCl and mass of CaCl2 in the original sample. State the mass of each salt as: (mole of salt) x (gram formula weight of salt). (c) Write the algebraic equation that relates the mass of the sample to the mass of (d) Calculate the percent CaCl, in the original sample. %CaCl2 = (e) Calculate the percent NaCl in the original sample. % NaCl_

Explanation / Answer

a)

NaCl (aq) + AgNO3 (aq) ------> AgCl(s) + NaNO3 (aq)

2 AgNO3 (aq) + CaCl2 (aq) ------> 2 AgCl (aq) + Ca(NO3)2 (aq)

Final Equation is

NaCl(aq) + CaCl2(aq) + 3 AgNO3(aq) ------> 3 AgCl (s) + NaNO3(aq) + Ca(NO3)2(aq)

b)

1 mole of NaCl + 1 mole of CaCl2 + 3 moles of AgNO3 ------> 3 moles of AgCl + 1 mole of NaNO3 + 1 mole of Ca(NO3)2

1 mole of NaCl produces 1 mole of AgCl

1 mole of CaCl2 produces 2 moles of AgCl

c)

1 mole * 58.5 g/mole of NaCl produces 1 mole * 143.4 g/mole of AgCl

1 mole * 111.1 g/mole of CaCl2 produces 2 moles * 143.4 g/mole of AgCl

the mixture contain both NaCl and CaCl2 then,

(58.5 + 111.1) g of NaCl+CaCl2 mixture produces 143.4 +(2*143.4) g of AgCl

169.6 g of mixture produces 430.2 g of AgCl

so 430.2 g of AgCl requires 111.1 g of CaCl2 solution

12.554 g of AgCl requires 111.1 * 12.554/430.2 g of CaCl2 solution

                         = 3.2421 g of CaCl2 solution

percetage of CaCl2 solution = mass of CaCl2/total mass of mixture*100

                            = 3.2421/5*100 = 64.84% of CaCl2 is present.

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