Wednesday
October 1, 2014

Homework Help: Chemistry

Posted by Aparicio on Wednesday, June 10, 2009 at 3:54pm.

A reaction was run where lead nitrate and potassium iodide were mixed. The resulting precipitate only gave a 78.1% yield. If the actual yield was 25.0grams, how many grams of lead nitrate were used?

This is what I have so far:

Potassium Iodide= KI = 166.0028g/mol
lead nitrate= Pb(NO3)2 =331.2 g/mol

I do not see how I can obtain the grams of nitrate from the information that is given can someone please help me?

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