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
(a) ___KOH + ___H3PO4 → ___K3PO4 + ___H2O
H3po4 + 3koh ----> k3po4 + 3h2o
P4o10 + 6h2o ----> 4h3po4
The numbers of sodium, oxygen, and hydrogen atoms are the same on both sides of the equation.
H3PO4 + 3NaOH --------> Na3PO4 + 3H2O
There are two reaction equations for phosphate ion and water. They two equations are: i) HPO4 2- + H2O => H3O+ + PO4 3- ii) HPO4 2- + H2O => OH- + H2PO4 -
H3PO4 --------> H+ + H2PO4- + H+ ----------> HPO42- + H+
The chemical equation is:3 NaOH + H3PO4 = Na3PO4 + 3 H2O
my balance sheet does not balance why?
The numbers of sodium, oxygen, and hydrogen atoms are the same on both sides of the equation.
Na3PO4+H2O->NaOH+H3PO4 just balance it.
Balanced equation:12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7
The balanced equation is 3 Ca(OH)2 + 2 H3PO4 -> Ca3(PO4)2 + 6 H2O.
H3PO4 + 3NaOH --------> Na3PO4 + 3H2O
There are two reaction equations for phosphate ion and water. They two equations are: i) HPO4 2- + H2O => H3O+ + PO4 3- ii) HPO4 2- + H2O => OH- + H2PO4 -
H3PO4 --------> H+ + H2PO4- + H+ ----------> HPO42- + H+
The chemical equation is:3 NaOH + H3PO4 = Na3PO4 + 3 H2O
To solve this problem, you will need to use the balanced chemical equation provided to determine the mole ratios between the reactants and products. First, convert the volume of H3PO4 to liters by dividing by 1000 mL/L: 750 mL H3PO4 / 1000 mL/L = 0.750 L H3PO4 Next, convert the concentration of H3PO4 to moles/L: 6.00 M H3PO4 = 6.00 mol/L H3PO4 Now, use the volume and concentration to calculate the number of moles of H3PO4: 0.750 L H3PO4 * 6.00 mol/L H3PO4 = 4.50 mol H3PO4 Since the chemical equation shows a 1:1 mole ratio between H3PO4 and Ca(OH)2, there must be 4.50 mol Ca(OH)2 as well. To determine the mass of each product, you will need to know the molar masses of each compound. The molar mass of H3PO4 is 98.00 g/mol, and the molar mass of Ca(PO4)2 is 212.09 g/mol. Therefore, the mass of H3PO4 produced in the reaction is: 4.50 mol H3PO4 * 98.00 g/mol = 434.00 g H3PO4 And the mass of Ca(PO4)2 produced in the reaction is: 4.50 mol Ca(PO4)2 * 212.09 g/mol = 953.41 g Ca(PO4)2 These are the masses of each product that would be produced if 750 mL of 6.00 M H3PO4 reacts according to the given chemical equation.
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
H3PO4 (aq) + LiOH (aq) ---> H2O(liq) + Li3PO4 (aq) Second opinion: That was pretty close, but Li and H are not balanced. Try this: H3PO4 (aq) + 3 LiOH (aq) ---> 3 H2O (liq) + Li3PO4 (aq)