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How do you balance this equation?

How do you balance this equation? Topic: Third grade homework assignments
June 25, 2019 / By Audra
Question: PO₄ + H₂O → H₃PO₄ This problem was on a recent homework assignment in my chemistry class, and NO ONE could figure it out. We're only in 8th grade Honors science, so no answers that are super-complicated, if possible. And if you don't know how to type subscripts you can just copy-paste, or just use the regular number, I'll understand.
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Best Answers: How do you balance this equation?

Abbey Abbey | 6 days ago
This not a chemistry equation nor a math question. As it is, it cannot be balanced chemically PO4 is a radical and should be written as PO4^3-.. You can end up with something that is balanced , but it is not chemistry PO4. + H2O --> H3PO4 To balance the hydrogen you have to multply the water by three and the H3PO4 by two: . Like so: 3H2O 2H3PO4 The PO4 would have to have a 2 to give you 2 PO4 + 3H2O --> 2 H3PO4. you can add 3 O.(ugh) Try these for practice: NaOH + H2SO4 --> Na2SO4 + H2O, and AgNO3 + CaCl2 --> AgCl + Ca(NO3)2
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Abbey Originally Answered: Correct Balance Equation?
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.
Abbey Originally Answered: Correct Balance Equation?
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)
Abbey Originally Answered: Correct Balance Equation?
NO in first case sure in 2nd just ensure that the numbers of each and every element is the equal on either side of the equation 4Fe(s) + 3O2(g) ----> 2Fe2O3(s) 2NaHCO3(s) ----> Na2O(s) + H2O(l) + 2CO2(g) despite the fact that there are intermediate levels to this response 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) Na2CO3(s) → Na2O(s) + CO2(g)

Spirit Spirit
The equation you wrote is missing a couple of things, like the charge on the phosphate ion PO₄ should be -3. PO₄ + 3H₂O → H₃PO₄ + 3OH I don't know how to write superscripts, but the PO₄ on the left side should have a "3-" as the superscript and the OH on the right should have a "-" as the superscript. That would balance the charges and the elements on both sides of the equation.
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Spirit 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!
Spirit Originally Answered: How do I balance this net ionic equation?
Na(+) + 2CO(-) + Ca(2+) + 2CL(-) --> CaCO3(s) + Na(+) + Cl(-) Ag(2+) + 2NO3(-) + ok(2+) + I(2-) --> AgI(s) + ok(2+) + 2NO3(-) 2H(+) + SO4(2-) + Na(+) + 2OH(-) --> 2H2O + 2Na(+) + SO4(2-) internet ionic equations pass away out the areas of the equation that have do no longer pass into making the precipitate. i'm going to assist you to separate those out from the above equations. The precipitate (or water) is on no account taken aside on the proper factor of the equation by fact it relatively isn't any longer an element of the reaction.

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