Look at the products. You should already know that one is bound to be water, H2O. You know that the OH comes from KOH, and the H comes from H3PO4. So ask yourself the question: How many OH and how many H are available? You can see that since you have three of one, you will also need three of the other, so you need 3 of the KOH, right? So this is the equation:
3KOH + H3PO4 -> 3H2O + K3PO4
To balance the reaction between H3PO4 and NaOH, first write out the chemical equation: H3PO4 + 3 NaOH -> Na3PO4 + 3 H2O Now, each element must have the same number of atoms on both sides of the equation for it to be balanced. Adjust the coefficients in front of each compound to achieve this balance.
H3PO4 is Phosphoric Acid. What are you reacting it with, in order to balance the eq'n.
K3PO4 + 3HCl -> 3KCl + H3PO4 Since K on the reactant side has 3 potassium atoms, K on the product side should also have 3 potassium atoms to balance the equation. Since you put the 3 on KCl of the product side, another 3 has to go on the Cl on the reactant side which also matches the 3 hydrogen atoms on the product side in H3PO4. If you check, the equation is now balanced. Everything that appears on the left, equally appears on the right
To demonstrate the conservation of mass in the chemical reaction 3NaOH + H3PO4 -> Na3PO4 + 3H2O, you would need to show that the total mass of the reactants (3NaOH + H3PO4) is equal to the total mass of the products (Na3PO4 + 3H2O). This can be done by calculating the total mass of each side of the equation using the molar masses of the compounds and ensuring they are equal. This illustrates that mass is conserved in the reaction.
The dissociation of 2 protons from trihydrogen phosphate (H3PO4) results in the formation of dihydrogen phosphate (H2PO4-) and a water molecule (H2O). The chemical equation for this process can be written as: H3PO4 ⇌ H2PO4- + H+ + H2O
To balance the reaction between H3PO4 and NaOH, first write out the chemical equation: H3PO4 + 3 NaOH -> Na3PO4 + 3 H2O Now, each element must have the same number of atoms on both sides of the equation for it to be balanced. Adjust the coefficients in front of each compound to achieve this balance.
H3PO4 is Phosphoric Acid. What are you reacting it with, in order to balance the eq'n.
my balance sheet does not balance why?
The chemical equation for phosphoric acid is H3PO4.
Balanced equation:12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7
K3PO4 + 3HCl -> 3KCl + H3PO4 Since K on the reactant side has 3 potassium atoms, K on the product side should also have 3 potassium atoms to balance the equation. Since you put the 3 on KCl of the product side, another 3 has to go on the Cl on the reactant side which also matches the 3 hydrogen atoms on the product side in H3PO4. If you check, the equation is now balanced. Everything that appears on the left, equally appears on the right
Na3PO4+H2O->NaOH+H3PO4 just balance it.
The dissociation of 2 protons from trihydrogen phosphate (H3PO4) results in the formation of dihydrogen phosphate (H2PO4-) and a water molecule (H2O). The chemical equation for this process can be written as: H3PO4 ⇌ H2PO4- + H+ + H2O
To demonstrate the conservation of mass in the chemical reaction 3NaOH + H3PO4 -> Na3PO4 + 3H2O, you would need to show that the total mass of the reactants (3NaOH + H3PO4) is equal to the total mass of the products (Na3PO4 + 3H2O). This can be done by calculating the total mass of each side of the equation using the molar masses of the compounds and ensuring they are equal. This illustrates that mass is conserved in the reaction.
The chemical equation is:3 NaOH + H3PO4 = Na3PO4 + 3 H2O
The balanced chemical equation for the reaction between cesium hydroxide solution (CsOH) and phosphoric acid (H3PO4) is: 3 CsOH + H3PO4 → Cs3PO4 + 3 H2O This equation is balanced because there is an equal number of atoms of each element on both sides of the equation.
When solving this type of problem, first use the ion charges to predict the formulas of the products. Then use coefficients to balance the equation. H3PO4 (aq) + 3 KOH (aq) --> K3PO4 (aq) + 3 H2O (l)