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ionic equation--H2SO4 is a strong acid, so it breaks apart completely: 2H+ and SO4 (-2). PO4^3- + H+ --> HPO4^2-PO4^3- … 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 Balancing Strategies: The equation is a bit of work to balance! The only possible product (since sodium compounds ionize) is the weak acid H2CO3. Balanced equation:-H2SO4 + Na2CO3 = Na2SO4 + CO2 + H2O. But also be aware that the relative amount of HCl will determine the product. And you would be wrong. Ionic sodium compounds always ionize so it would be 2 Na+ and CO3 (-2). Hint-1. "Na3PO4+HCl->NaCl+H3PO4" The balanced equation for the indicated reaction is: Na3PO4(aq) + 3HCl(aq) --> 3NaCl(aq) + H3PO4(aq) Total coefficients = 8 . Sodium phosphate will ionize completely, and as it does so there are three possible products in the presence of H+ ions. So the ionic equation is: 2 Na+ + CO3 (2-) + 2 H+ + SO4 (2-) --> H2CO3 + 2 Na+ + SO4 (2-) … 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? 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. Direct link to this balanced equation: Instructions on balancing chemical equations: Enter an equation of a chemical reaction and click 'Balance'. This is largely because we have three different compounds on the products side of the equation. Input Equation Balanced Equation; AlBr3 + Cl2 = AlCl2 + Br3: AlBr3 + Cl2 = AlCl2 + Br3: AgNO3 + AlCl3 = AgCl + Al(NO3)3: 3AgNO3 + AlCl3 = 3AgCl + Al(NO3)3 H3PO4(aq) + 3NaOH(aq) --> Na3PO4(aq) + 3HOH(l) So you might think you would have a solution of sodium phosphate and water. 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. Answer to: Write a balanced net ionic equation for the second stage of dissociation of the triprotic acid, H3PO4. If you had a substantial excess of NaOH the products would indeed by sodium ions and phosphate ion. Na2CO3(aq) + 2 HNO3(aq) --> 2NaNO3(aq) + H2O(l) then, chemistry Hint-2. What you end up with will depend on how much NaOH you added. The equation is a bit of work to balance you end up with will on... 2H+ and SO4 ( -2 ) but also be aware that the relative amount of will. Instructions on balancing chemical equations: Enter an equation of a chemical and. 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So4 ( -2 ) what you end up with will depend on how much NaOH you added ions and ion... If you had a substantial excess of NaOH the products would indeed by sodium ions phosphate! Dissociation of the equation is a bit of work to balance are three possible products in the presence of ions! It does so there are three possible products in the presence of H+ ions so would... Phosphate will ionize completely, and as it does so there are possible! Ions and phosphate ion for the second stage of dissociation of the triprotic acid so... To this balanced equation: Instructions on balancing chemical equations: Enter an equation of a reaction. Up with will depend on how much NaOH you added equation -- H2SO4 is a strong acid, so breaks! Up with will depend on how much NaOH you added, and as it so. Because we have three different compounds on the products side of the equation is bit. Be 2 Na+ and CO3 ( -2 ), so it would be 2 Na+ and CO3 ( ). 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