The answer will appear below; Always use the upper case for the first character in the element name and the lower case for the second character. PO4^3- + H+ --> HPO4^2-PO4^3- … The only possible product (since sodium compounds ionize) is the weak acid H2CO3. Direct link to this balanced equation: Instructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. Balancing Strategies: The equation is a bit of work to balance! This is largely because we have three different compounds on the products side of the equation. "Na3PO4+HCl->NaCl+H3PO4" The balanced equation for the indicated reaction is: Na3PO4(aq) + 3HCl(aq) --> 3NaCl(aq) + H3PO4(aq) Total coefficients = 8 . But also be aware that the relative amount of HCl will determine the product. ionic equation--H2SO4 is a strong acid, so it breaks apart completely: 2H+ and SO4 (-2). Na2CO3(aq) + 2 HNO3(aq) --> 2NaNO3(aq) + H2O(l) then, chemistry What you end up with will depend on how much NaOH you added. H3PO4(aq) + 3NaOH(aq) --> Na3PO4(aq) + 3HOH(l) So you might think you would have a solution of sodium phosphate and water. If you had a substantial excess of NaOH the products would indeed by sodium ions and phosphate ion. Balanced equation:-H2SO4 + Na2CO3 = Na2SO4 + CO2 + H2O. Ionic sodium compounds always ionize so it would be 2 Na+ and CO3 (-2). Hint-1. Hint-2. Answer to: Write a balanced net ionic equation for the second stage of dissociation of the triprotic acid, H3PO4. So the ionic equation is: 2 Na+ + CO3 (2-) + 2 H+ + SO4 (2-) --> H2CO3 + 2 Na+ + SO4 (2-) … I recommend you start by getting an even number of Na atoms by changing the coefficient on the Na3PO4 (multiply by two to get an even number so you can change the Na2CO3 to balance the Na atoms). Despite two other answers giving you clear, unambiguous answers (and agreeing on the answer), this question can’t be answered without more information. Sodium phosphate will ionize completely, and as it does so there are three possible products in the presence of H+ ions. And you would be wrong. what volume (in ML) of a 0.150 M HNO3 solution is required to completely react wtih 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation? Input Equation Balanced Equation; 4C6H5F+O2=CO2+10H2O+2F2: 4C6H5F + 29O2 = 24CO2 + 10H2O + 2F2: 4P+3O2=2P2O3: 4P + 3O2 = 2P2O3: 4P+3O2=2P2O3: 4P + 3O2 = 2P2O3 Input Equation Balanced Equation; AlBr3 + Cl2 = AlCl2 + Br3: AlBr3 + Cl2 = AlCl2 + Br3: AgNO3 + AlCl3 = AgCl + Al(NO3)3: 3AgNO3 + AlCl3 = 3AgCl + Al(NO3)3
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