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How do I balance this equation?

Topic: How to write a net ionic equation for a reaction
June 25, 2019 / By Teri
Question: Write a balanced chemical equation for the reaction of copper (II) sulfate and concentrated ammonia to produce tetraamminecopper (II) sulfate. Is this correct? CuSO4 + NH3 -> Cu(NH3)4SO4 Help me balance it??

Best Answers: How do I balance this equation?

Rosa | 2 days ago
Molecular equation: CuSO4(aq) + 4NH3(aq) --> [Cu(NH3)4]SO4(aq) Net ionic equation: Cu2+ + 4NH3(aq) --> [Cu(NH3)4]^2+ ...... SO4^2- is a spectator ion When ammonia solution is added to a solution of copper ions, you will first see a gelatinous blue solid forming. It consists of insoluble Cu(OH)2. The OH- comes from the basic solution of ammonia. As more NH3 is added, the precipitate dissolves as more [Cu(NH3)4]^2+ is produced.
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We found more questions related to the topic: How to write a net ionic equation for a reaction

You wrote, "Cu(NO3)2 (aq) + NH4OH (aq) -----> CuOH (s) NH4(NO3)2 (aq)" This equation is a nonstarter. The equation should look like this:: Cu(NO3)2(aq) + 2NH4OH(aq) ---> Cu(OH)2(s) + 2NH4NO3(aq) There is no CuOH formed and there is no such compound as NH4(NO3)2. The next problem is that there is no NH4OH. What we call ammonium hydroxide is actuall ammonia (NH3) dissolved in water. A more accurate equation will look like this: Cu(NO3)2(aq) + 2NH3(aq) + 2H2O(l) ---> Cu(OH)2(s) + 2NH4NO3(aq) The net ionic equation for the first version... Cu(NO3)2(aq) + 2NH4OH(aq) ---> Cu(OH)2(s) + 2NH4NO3(aq) ... the net ionic equation is: Cu2+ + 2OH- --> Cu(OH)2(s) ...... NH4+ and NO3- are spectator ions For the more accurate equation the net ionic equation is: Cu2+ + 2NH3(aq) + 2H2O(l) ---> Cu(OH)2(s) + 2NH4+ ============ Follow up ============== With all due respect, you do NOT have it written correctly, and neither gp4rts nor Stefany know what they are talking about. As I said above, "There is no CuOH formed and there is no such compound as NH4(NO3)2." =========== More follow up ============ For one thing, you need to pay attention to the oxidation numbers of the elements and polyatomic ions that you are putting together to form the formula of the products. For instance, in double replacement reactions there are no oxidation numbers changes (redox), so be sure adjust the subscripts so that the sum of the oxidation numbers is zero. Cu == +2 and OH == -1. Therefore, you need two OH- for each Cu.... Cu(OH)2.... +2 + 2(-1) = 0 NH4 == +1, NO3 == -1. They combine in a 1:1 ratio to give an oxidation number sum of zero. Don't bring the subscripts to the products from the reactants. The subscripts of the products must be determined from the oxidation numbers of the species you are combining. Water is often a product in acid/base reactions. It is also formed when carbonates or bicarbonates react with acids. Water will always be a product in the complete combustion of a hydrocarbon. Acid/base: 2HC2H2O2(aq) + Ca(OH)2(s) --> Ca(C2H2O2)2(aq) + 2H2O(l) Acid/carbonate: HCl(aq) + NaHCO3(s) --> NaCl(aq) + CO2(g) + H2O(l) Combustion: C5H12(g) + 8O2(g) --> 5CO2(g) + 6H2O(l) Using a solubility table is a **good** thing. Continue to include the state symbols in the equation. It will give you a better idea of how all this stuff looks, like what is dissolved in water, what is precipitating out, and what just wandered off as a gas. I require my students to include state symbols when they write chemical equations and particularly when they are predicting the products.
You have written it correctly but the formula is incorrect. The formula for copper hydroxide is Cu(OH)2 and for ammonium nitrate NH4NO3 , so the reaction should be Cu(NO3)2(aq) + 2NH4OH(aq) ---------> Cu(OH)2 + 2NH4NO3(aq)

Monta
This Site Might Help You. RE: How do I balance this equation? Write a balanced chemical equation for the reaction of copper (II) sulfate and concentrated ammonia to produce tetraamminecopper (II) sulfate. Is this correct? CuSO4 + NH3 -> Cu(NH3)4SO4 Help me balance it??
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Lisanne
For the best answers, search on this site https://shorturl.im/1O3wT Assuming you really mean carbon monoxide and not dioxide then: TiO2 (s) + 2C (s) + 2Cl2(g) TiCl4(s?) + 2CO(g) Same number of each type of atom on each side. Start by looking at the highest number in this case 4 inTiCl4, it comes from Cl2 so need 2Cl2. Then O start with 2 so must end with 2 so 2CO so need 2C on the other side to provide the other carbon. if you are producing carbon dioxide then: TiO2 + C + 2Cl2 TiCl4 + CO2 Sorry am not sure of physical state of TiCl4. (s) indicates solid, (g) indicates gas, (aq) in aqueous solution, (l) liquid.
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Originally Answered: How do I balance this net ionic equation?
Separate the redox equation into two reduction/oxidation equations: I3- ----> I- and S2O3 2- -----> S4O6 2- (This makes it easier to solve individually instead of doing everything at once). Starting with I3- ------>I - To balance the I, you need to put 3 as a coefficient in front of I- (right) Thus, the equation becomes: I3- ------> 3 I- However, the equation is not balanced: since the left side has 1 as a coefficient, the left side of the equation has 1 mole of electron. The right side has 3 as a coefficient, thus having 3 moles of electrons. Therefore, to balance, you need to add 2 electrons to the right equation. I3- + 2e- ------>3 I- Think of it this way: coefficient times charge equals moles of electrons. Thus, for the reaction S2O3 2- -----> S4O6 2-, first balance the number of atoms by adding 2 as a coefficient to the left side. 2 S2O3 2- ------> S4O6 2- The number of atoms are the same, but the mole of electrons are not...on the left is 4 mole electrons, and on the right is two mole electrons. Thus, you must add 2 mol electrons to the right side. 2 S2O2 2- ------>S4O6 2- + 2e- This problem doesn't require any multiplying as there is the same number of moles of electrons on opposite sides. Thus, analyzing the two equations. I3- + 2e- ----->3 I- 2 S2O3 2- -----> S4O6 2- +2e- Cancel out the electrons (since they appear on alternate sides of different equations). Add reactants and products to get: I3- + 2 S2O3 2- ----> S4O6 2- + 3 I- To check, multiple charges by coefficients. Reactants: 1(-1) +2(-2)--->-5 mole electrons Products: 1(2-) + 3(1-)---->5 mole electrons Thus, the equation is: I3- + 2 S2O3 2- ----> S4O6 2- + 3 I- Make sure you study hard for that AP test!

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