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3. (Problem 5) Each of the following metal chlorides can be reacted with silver nitrate to form a precipitate. For a 0.2500 g sample of the metal chloride, determine what mass of precipitate would be formed Show how the obtained mass of precipitate could be used to determine the identity of the metal by determining the percent by mass of chlorine in the compound D CuCl2

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Balanced chemical reaction CuCI2(aq) + 2 AgNO3(aq) 2AgCl(s) + Cu2+(aq) + 2NO3-(aq) molar mass of CuCI2 moles of CuCI2 sample 134.45 g/mol 0.2500g / 13445g/mol 0.001860 moles we use little excess of AgNO3 to completely precipitate Cl- present as AgCl insoluble salt moles of Agcl produced molar mass of Agc mass of Agcl produced -2*0.001860 moles 0.00372 moles -143.32 g/mol 0.00372mol 143.32g/mol 0.5332 g The mass % in AgCl %ofAg %Of Cl Thus in 0.5332 g of AgCl we will have mass of CI i.e. moles of CI mass of metal in the sample % mass of Cl in the sample mole of metal in sample atomic mass of metal Thus the metal with molar mass 63.49 is Cu and hence we can identify metal on the basis of mass of AgCl ppt 75.263% 24.737% 0.5332 0.247370.1319 g of Ci- 0.1319 /35.5 g/mol 0.00372 moles 0.25-0.1319 0.1181 g (0.1319/0.25)*100-52.76% 0.00372 2 0.00186 moles (0.1181g)/0.00186 mole 63.49 g/mol

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