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164.00 mL of 0.200 M Cu(NO3)2(aq) is mixed with 93.000 mL of 0.400 M NaOH(aq). Assuming 100% yield, what mass (in g) of precipitate will form? Provide your answer in decimal notation rounded to the appropriate number of significant figures.

164.00 mL of 0.200 M Cu(NO3)2(aq) is mixed with 93.000 mL of 0.400 M NaOH(aq). Assuming 100% yield, what mass (in g) of precipitate will form? Provide your answer in decimal notation rounded to the appropriate number of significant figures.

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem ISP: Aqueous solutions of ammonium sulfide and mercury(II) nitrate react and a precipitate forms. (a)...
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164.00 mL of 0.200 M Cu(NO3)2(aq) is mixed with 93.000 mL of 0.400 M NaOH(aq). Assuming 100% yield, what mass (in g) of precipitate will form? Provide your answer in decimal notation rounded to the appropriate number of significant figures.
Transcribed Image Text:164.00 mL of 0.200 M Cu(NO3)2(aq) is mixed with 93.000 mL of 0.400 M NaOH(aq). Assuming 100% yield, what mass (in g) of precipitate will form? Provide your answer in decimal notation rounded to the appropriate number of significant figures.
Expert Solution
Step 1

The chemical reaction is:

Cu(NO3)2 (aq) + 2NaOH (aq) →Cu(OH)2 (s) + 2NaNO3 (aq)

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