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Excess silver nitrate is reacted with 0.1442 g unknown chloride forming a white precipitate. The mass of the precipitate is 0.1194 g. Calculate the percent by mass of chloride in the unknown solid.
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I'll use X to represent the unknown cation in the unknown chloride. going by solubility rules, the precipitate is AgCl, not the nitrate. AgNO3 + XCl --> AgCl + XNO3 use stoich to convert from g AgCl --> moles AgCl --> moles Cl --> grams Cl now, this gives you the total mass of chloride in the original sample. after that it's simply (mass of chloride)/(total mass of unknown chloride salt) * 100%
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