A 10.000 gm sample of a mixture containing aluminum chloride was titrated with a
ID: 699063 • Letter: A
Question
A 10.000 gm sample of a mixture containing aluminum chloride was titrated with a solution of silver nitrate. The silver nitrate solution was prepared by adding 10.000 gm of silver nitrate to enough water to a volume of 250 ml. If the aluminum Chloride solution required 27.96 ml of the silver nitrate solution, what is the % of aluminum chloride in the sample? A 10.000 gm sample of a mixture containing aluminum chloride was titrated with a solution of silver nitrate. The silver nitrate solution was prepared by adding 10.000 gm of silver nitrate to enough water to a volume of 250 ml. If the aluminum Chloride solution required 27.96 ml of the silver nitrate solution, what is the % of aluminum chloride in the sample? Chloride solution required 27.96 ml of the silver nitrate solution, what is the % of aluminum chloride in the sample?Explanation / Answer
Balanced equation
3AgNO3(aq) + AlCl3(aq) ====> Al(NO3)3(aq) + 3AgCl(s)
Reaction type: double replacement
Molarity of the Silver nitrate = 10 x 250 / (1000 x 169.87) = 0.23457M
Moles of silver nitrate used in the titration = 0.23457 x 27.96 /1000 = 0.00658 Moles
Moels of AlCl3 present in the sample = 0.00219
Mass of the AlCl3 = 0.00219 x 133.34 = 0.2920 gm
% of aluminum chloride in the sample = 0.2920 x 100 / 10 = 2.92 %
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