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Treatment of a 0.2500-g sample of impure potassium chloride with an excess of AgNO3 resulted in the for- mation of 0.2912 g of AgCl. Calculate the percentage of KCl in the sample.

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2 answers

  1. 0.0020315*74.5513 = 0.1513
    % = 0.1513/0.2500 * 100 =60.52

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  2. KCl + AgNO3 ==> AgCl + KNO3
    mols AgCl formed = g/molar mass = 0.2912/143.34 = 0.0020315
    The ratio of KCl to AgCl is 1:1; therefore, mols KCl to start must be 0.0020315. Grams KCl = mols KCl x molar mass KCl = ?
    % KCl = (grams KCl/mass sample)*100 = ?
    Post your work if you get stuck. Check those molar mass and calculations. I tried to do that from memory.

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