For the reaction 3KOH+H3PO4⟶K3PO4+3H2O

how many grams of phosphoric acid, H3PO4, are needed to react completely with 89.7 g of potassium hydroxide, KOH?

52.2 g

Explanation:

Step 1: Write the balanced equation

3 KOH + H₃PO₄ ⟶ K₃PO₄ + 3 H₂O

Step 2: Calculate the moles corresponding to 89.7 g of KOH

The molar mass of KOH is 56.11 g/mol.

89.7 g × 1 mol/56.11 g = 1.60 mol

Step 3: Calculate the moles of H₃PO₄ needed to react with 1.60 moles of KOH

The molar ratio of KOH to H₃PO₄ is 3:1. The moles of H₃PO₄ needed are 1/3 × 1.60 mol = 0.533 mol.

Step 4: Calculate the mass corresponding to 0.533 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.533 mol × 97.99 g/mol = 52.2 g

1 a

2 d

3 a

4 c

explanation:

your answer would be “dry”