1. In an experiment, suppose 10.0 g of potassium iodide was used with excess lead (II) nitrate & 9.50 g of precipitate was collected. Calculate the percentage yield of the reaction.(I already know how to do this but just for reference for second question)
2.In the previous example, 10.0 g of potassium iodide was reacted with lead (II) nitrate. What if the potassium iodide was only 85.0% pure? How much precipitate would we expect then?
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Answer:
Hii there.
Explanation:
In this double displacement precipitation reaction, lead II nitrate solution is added to potassium iodide solution. Iodide treatment of water can be used as an indicative test for the presence of the lead II ion.
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