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This equation is P4O10 + 6 H2O -> 4 H3PO4.
H3PO4 + 3NaOH --------> Na3PO4 + 3H2O
3CsOH(aq) + H3PO4(aq) --> Cs3PO4(aq) + 3H2O(l)
Usually these reaction are done with a substantial excess of NaOH. You would getH3PO4 + 3NaOH --> Na3PO4(aq) + 3HOH.
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
Balanced equation:12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7
This equation is P4O10 + 6 H2O -> 4 H3PO4.
H3PO4 + 3NaOH --------> Na3PO4 + 3H2O
3CsOH(aq) + H3PO4(aq) --> Cs3PO4(aq) + 3H2O(l)
H3PO4 + NaOH ----> Na2HPO4 + H2O
The balanced equation is 3 Ca(OH)2 + 2 H3PO4 -> Ca3(PO4)2 + 6 H2O.
Usually these reaction are done with a substantial excess of NaOH. You would getH3PO4 + 3NaOH --> Na3PO4(aq) + 3HOH.
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
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)
it may produces explosive material if not stabilized
Your reaction is this...H3PO4 + 3KOH --> K3PO4 + 3H2OThis type of reaction is called a double replacement.Note: You didn't put a 4 on the oxygen on the phosphate group on the product side in your question. If it wasnt there the equation would not have been balanced so i put it in myself.