Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.99 g of precipitate. Write the balanced molecular, total ionic, and net ionic equations of the reaction. Atomic Mass: Pb: 207.2 g/mol N: 14.0067 g/mol O: 15.999 g/mol Na: 22.989769 g/mol I: 126.90447 g/mol Note: Use scientific notation with the format (n)e(x) where n = number and x = exponent (Example: 6.022e23 for 6.022 x 1023) Round your answer to 4 decimal places.

Chemistry: Principles and Reactions
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Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter4: Reactions In Aqueous Solution
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Question 9
O Points
Lead (II) nitrate is reacted with sodium iodide. Calculate
the number of Pb2+ ions required to form 0.99 g of
precipitate. Write the balanced molecular, total ionic, and
net ionic equations of the reaction.
Atomic Mass:
Pb: 207.2 g/mol
N: 14.0067 g/mol
O: 15.999 g/mol
Na: 22.989769 g/mol
I: 126.90447 g/mol
Note: Use scientific notation with the format (n)e(x) where
n = number and x = exponent (Example: 6.022e23 for
6.022 x 1023)
Round your answer to 4 decimal places.
Transcribed Image Text:Question 9 O Points Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.99 g of precipitate. Write the balanced molecular, total ionic, and net ionic equations of the reaction. Atomic Mass: Pb: 207.2 g/mol N: 14.0067 g/mol O: 15.999 g/mol Na: 22.989769 g/mol I: 126.90447 g/mol Note: Use scientific notation with the format (n)e(x) where n = number and x = exponent (Example: 6.022e23 for 6.022 x 1023) Round your answer to 4 decimal places.
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