for the following?

Molar mass of KOH = (39.01 + 16.00 + 1.008) g/mol = 56.02 g/mol
Molar mass of H₃PO₄ = (1.008×3 + 30.97 + 16.00×4) g/mol = 97.99 g/mol
Molar mass of K₃PO₄ = (39.01×3 + 30.97 + 16.00×4) g/mol = 212.0 g/mol

Initial number of moles of KOH = (67.0 g) / (56.02 g/mol) = 1.196 mol
Initial number of moles of H₃PO₄ = (43.1 g) / (97.99 g/mol) = 0.4398 mol

3KOH(aq) + H₃PO₄(aq) → K₃PO₄(aq) + 3H₂O(l)
Mole ratio KOH : H₃PO₄ = 3 : 1

If KOH completely reacts :
Number of moles of H₃PO₄ needed = (1.196 mol) × (1/3) = 0.3987 mol < 0.4398 mol
Hence, H₃PO₄ is in excess, and the limiting reagent is KOH.

According to the equation, mole ratio KOH : K₃PO₄ = 3 : 1
Number of moles of KOH reacted = 1.196 mol
Maximum number of moles of K₃PO₄ formed = (1.196 mol) × (1/3)
Maximum mass of K₃PO₄ formed = (1.196 mol) × (1/3) × (212.0 g/mol) = 84.5 g

Formula of the limiting reagent : KOH

In the calculations above, number of moles of H₃PO₄ reacted = 0.3987 mol
The number of moles of excess reagent (H₃PO₄) remains = (0.4398 - 0.3987) mol = 0.0411 mol
The mass of excess reagent (H₃PO₄) remains = (0.0411 mol) × (97.99 g/mol) = 4.03 g

The molar mass of KOH is 56.11 g/mol; that of H₃PO₄ is 97.99 g/mol.
Therefore we have (67.0 g)/(56.11 g/mol) = 1.19 mol KOH and (43.1 g)/(97.99 g/mol) = 0.440 mol H₃PO₄.

The reaction equation is 3KOH (aq) + H₃PO₄ (aq) → K₃PO₄ (aq) + 3H₂O (l).
3 mol of KOH are needed for each mol of H₃PO₄, so we need 3(0.440 mol) = 1.32 mol KOH for the given amount of acid. Since there are only 1.19 mol KOH available, the KOH is the limiting reagent.

The mole ratio of KOH to K₃PO₄ is 3:1, so 1/3 of 1.19 mol of K₃PO₄ can be formed: 0.397 mol.
The molar mass of K₃PO₄ is 212.27 g/mol, so we get (0.397 mol)(212.27 g/mol) = 84.3 g.

The mole ratio of H₃PO₄ to K₃PO₄ is 1:1, so 0.397 mol H₃PO₄ is expended.
This is (0.397 mol)(97.99 g/mol) = 38.9 g, which means that 43.1 g - 38.9 g = 4.2 g remain.

(All results here are limited to 3 significant figures, so some differences will occur due to rounding.)