Question

0.76 g of lead (II) nitrate was dissolved in a 50.00 mL of water and treated with 25.00 mL of 0.2010 M sodium sulfate in order to determine a content of lead (II) ion. A white precipitate was formed and it was collected and dried. Calculate the amount of the precipitate formed in this reaction.

Answer 1

Answer:

Explanation:

mass of reactants = mass of products= 75.76