H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)

How many grams of potassium hydroxide are required to completely neutralize 3.23 grams of phosphoric acid?

H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)

mols H3PO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols H3PO4 to mols KOH.
Now convert mols KOH to grams. g = mols x molar mass = ?

To determine the grams of potassium hydroxide (KOH) required to neutralize 3.23 grams of phosphoric acid (H3PO4), you need to use the stoichiometry of the balanced chemical equation.

The balanced equation is:
H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)

From the equation, you can see that one mole of H3PO4 reacts with three moles of KOH.

1 mole H3PO4 (molar mass: 98.00 g/mol) reacts with 3 moles KOH (molar mass: 56.11 g/mol)

Now you can calculate the moles of H3PO4 using the given mass:
moles H3PO4 = mass / molar mass
moles H3PO4 = 3.23 g / 98.00 g/mol = 0.033 moles

Since there is a 1:3 stoichiometric ratio between H3PO4 and KOH, we can calculate the moles of KOH required:
moles KOH = moles H3PO4 × (3 moles KOH / 1 mole H3PO4)
moles KOH = 0.033 moles × (3 moles KOH / 1 mole H3PO4) = 0.099 moles

Finally, to find the grams of KOH required, use the moles of KOH and its molar mass:
mass KOH = moles KOH × molar mass
mass KOH = 0.099 moles × 56.11 g/mol = 5.54 grams

Therefore, you would need approximately 5.54 grams of potassium hydroxide (KOH) to completely neutralize 3.23 grams of phosphoric acid (H3PO4).

To determine the number of grams of potassium hydroxide (KOH) required to completely neutralize 3.23 grams of phosphoric acid (H3PO4), you need to use the balanced chemical equation and the molar ratios between the two compounds.

1. Start by writing out the balanced chemical equation:
H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)

2. Determine the molar mass of phosphoric acid (H3PO4) and potassium hydroxide (KOH). These can be found on the periodic table:
Molar mass of H3PO4 = 3(1.01 g/mol) + 1(31.00 g/mol) + 4(16.00 g/mol) = 98.00 g/mol
Molar mass of KOH = 1(39.10 g/mol) + 1(16.00 g/mol) = 56.10 g/mol

3. Convert the given mass of phosphoric acid (H3PO4) into moles using its molar mass:
Moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4 = 3.23 g / 98.00 g/mol = 0.0330 mol

4. Use the balanced chemical equation to determine the molar ratio between phosphoric acid (H3PO4) and potassium hydroxide (KOH).
According to the equation, it takes 3 moles of KOH to react with 1 mole of H3PO4.

5. Calculate the number of moles of KOH required:
Moles of KOH = Moles of H3PO4 * (3 moles of KOH / 1 mole of H3PO4) = 0.0330 mol * (3 mol KOH / 1 mol H3PO4) = 0.0990 mol

6. Convert the moles of KOH into grams using its molar mass:
Mass of KOH = Moles of KOH * Molar mass of KOH = 0.0990 mol * 56.10 g/mol = 5.543 g

Therefore, 5.543 grams of potassium hydroxide (KOH) are required to completely neutralize 3.23 grams of phosphoric acid (H3PO4).