What I would first is to find the amount of mols of H3PO4. Since you know that 1 M means there is 1 mol of solution in every 1 liter, you can figure out the moles of the solution. The solution is 6M, so that means that to find the moles of solute in the solution you multiply .750 (750 milliliters converted to liters) by 6. I think you did something similar, but .0075 L makes 7.5 milliliters. This will give you 4.5 mols of H3PO4.
Now you can use stoichiometry to find the grams of Ca3(PO4)2. The balanced equation states that it takes 2 mols of H3PO4 to get 1 mol of Ca3(PO4)2:
4.5 mol H3PO4 * 1 mol Ca3(PO4)2 / 2 mol H3PO4 = 2.25 mol Ca3(PO4)2
Find the molar mass of Ca3(PO4)2 to convert the mols to grams:
3*40 + 2*31 + 8*16 = 310 grams / mol
2.25 mol Ca3(PO4)2 * 310 grams Ca3(PO4)2 / 1 mol Ca3(PO4)2 = 697.5 grams Ca3(PO4)2
You can use the same process to find the mass of water.
4.5 mol H3PO4 * 6 mol H2O / 2 mol H3PO4 = 13.5 mol H2O
13.5 mol H2O * 18 g H2O / 1 mol H2O = 243 g H2O
Hope that helped! NB this is not my work just saw it on another website
H3PO4 + 3NaOH --------> Na3PO4 + 3H2O
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
FeBr2 + H3PO4 <--> Fe3(PO4)2 + HBr
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H2PO4-(aq)+H3O(aq) --> H3PO4(aq)+H2O(l)
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.
H3PO4 + 3NaOH --------> Na3PO4 + 3H2O
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 + NaOH ----> Na2HPO4 + H2O
FeBr2 + H3PO4 <--> Fe3(PO4)2 + HBr
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Balanced equation:12 HClO4 + P4O10 = 4 H3PO4 + 6 Cl2O7
Dissolve into water and it actually reacts with the water to form H3PO4, phosphoric acid.
H3po4+ 3nh4oh -> 3h2o + (nh4)3po4
Na3PO4+H2O->NaOH+H3PO4 just balance it.
It produces Sodium dihydrogen Phosphate and Hydrogen iodide: NaI + H3PO4 ----> NaH2PO4 + HI
H2PO4-(aq)+H3O(aq) --> H3PO4(aq)+H2O(l)