Copper(II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper(II) hydroxide. a.) Write the net ionic
equation for the reaction. b.) Calculate the maximum mass of copper(II) hydroxide that can be formed when 2.00 g of sodium hydroxide is added to 80.0 mL of 0.500M Cu(NO3)2 (aq).
What I initially got based on the question was Cu(NO3)2 + NaOH --> Cu(OH)2, but I'm not sure what to do next or if that equation is even correct to start with.
And how would that be applied to part b?
Thanks
M.9 Net Ionic Equations
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Jay Solanki 3A
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Re: M.9 Net Ionic Equations
Hello!
Precipitation reactions always result in double displacement. So the combination of copper (II) Nitrate and sodium hydroxide will produce copper (II) hydroxide and sodium nitrate (Be sure to balance first!!). Using solubility rules, we can determine that Cu(OH)2 is the precipitate, and therefore will be the only species in the reaction that will not break down into its constituent ions. The rest of the compounds ionize, since they are in an aqueous solution. After removing spectator ions from the equation, you should be left with Cu^2+ + 2 OH^- --> Cu(OH)2. Hope this helps!
Precipitation reactions always result in double displacement. So the combination of copper (II) Nitrate and sodium hydroxide will produce copper (II) hydroxide and sodium nitrate (Be sure to balance first!!). Using solubility rules, we can determine that Cu(OH)2 is the precipitate, and therefore will be the only species in the reaction that will not break down into its constituent ions. The rest of the compounds ionize, since they are in an aqueous solution. After removing spectator ions from the equation, you should be left with Cu^2+ + 2 OH^- --> Cu(OH)2. Hope this helps!
-
Jay Solanki 3A
- Posts: 137
- Joined: Wed Sep 30, 2020 9:59 pm
- Been upvoted: 1 time
Re: M.9 Net Ionic Equations
Hello!
a) Precipitation reactions always result in double displacement. So the combination of copper (II) Nitrate and sodium hydroxide will produce copper (II) hydroxide and sodium nitrate (Be sure to balance first!!). Using solubility rules, we can determine that Cu(OH)2 is the precipitate, and therefore will be the only species in the reaction that will not break down into its constituent ions. The rest of the compounds ionize, since they are in an aqueous solution. After removing spectator ions from the equation, you should be left with Cu^2+ + 2 OH^- --> Cu(OH)2 as the net ionic equation. Hope this helps!
b) This basically deals with stoichiometry of the reaction from a. One thing to keep in mind is that for stoichiometric conversions, use the COMPLETE chemical equation. That is, do not use the net ionic equation found in part a, but the complete equation as full compounds in which the compounds are not broken into their constituent ions. Since quantities of both reactants are given, you should perform stoichiometric conversions for both reactants, and find which is limiting. For the 2 g of sodium hydroxide, ratios to be used in order are the molar mass of Na(OH), the mole-mole ratio of Na(OH) to copper (II) hydroxide, and then the molar mass of Cu(OH)2 to find the grams of Cu(OH)2. The process is different for the Copper nitrate, where you would multiply the volume by the given concentration first to obtain the moles of copper (ii) nitrate in solution, then convert this value to grams of Cu(OH)2 using the mole-mole ratio of Cu(NO3)2 to Cu(OH)2 and then the molar mass of copper (II) hydroxide to obtain grams of copper (II) hydroxide. Which ever reactant results in less product would be limiting, and the value obtained would be considered the theoretical yield (or maximum mass). Hope this helps!
a) Precipitation reactions always result in double displacement. So the combination of copper (II) Nitrate and sodium hydroxide will produce copper (II) hydroxide and sodium nitrate (Be sure to balance first!!). Using solubility rules, we can determine that Cu(OH)2 is the precipitate, and therefore will be the only species in the reaction that will not break down into its constituent ions. The rest of the compounds ionize, since they are in an aqueous solution. After removing spectator ions from the equation, you should be left with Cu^2+ + 2 OH^- --> Cu(OH)2 as the net ionic equation. Hope this helps!
b) This basically deals with stoichiometry of the reaction from a. One thing to keep in mind is that for stoichiometric conversions, use the COMPLETE chemical equation. That is, do not use the net ionic equation found in part a, but the complete equation as full compounds in which the compounds are not broken into their constituent ions. Since quantities of both reactants are given, you should perform stoichiometric conversions for both reactants, and find which is limiting. For the 2 g of sodium hydroxide, ratios to be used in order are the molar mass of Na(OH), the mole-mole ratio of Na(OH) to copper (II) hydroxide, and then the molar mass of Cu(OH)2 to find the grams of Cu(OH)2. The process is different for the Copper nitrate, where you would multiply the volume by the given concentration first to obtain the moles of copper (ii) nitrate in solution, then convert this value to grams of Cu(OH)2 using the mole-mole ratio of Cu(NO3)2 to Cu(OH)2 and then the molar mass of copper (II) hydroxide to obtain grams of copper (II) hydroxide. Which ever reactant results in less product would be limiting, and the value obtained would be considered the theoretical yield (or maximum mass). Hope this helps!
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