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For the following reaction, 69.9 grams of potassium hydroxide are allowed to react with 37.9 grams of phosphoric acid:

\[ \text{potassium hydroxide (aq) + phosphoric acid (aq) } \rightarrow \text{ potassium phosphate (aq) + water (l)} \]

1. What is the maximum amount of potassium phosphate that can be formed?
2. What is the formula for the limiting reagent?
3. What amount of the excess reagent remains after the reaction is complete?

Answer :

Final answer:

The maximum amount of potassium phosphate that can be formed is 0.0975 moles. The formula for the limiting reagent is phosphoric acid (H3PO4). After the reaction is complete, there will be 1.40 moles of excess potassium hydroxide remaining.

Explanation:

To determine the maximum amount of potassium phosphate formed and the limiting reagent, we need to compare the moles of potassium hydroxide and phosphoric acid.

First, we calculate the moles of each reactant using their molar masses:

  • Potassium hydroxide (KOH): molar mass = 39.10 g/mol
  • Phosphoric acid (H3PO4): molar mass = 97.99 g/mol

Using the given masses:

  • Mass of potassium hydroxide = 69.9 g
  • Mass of phosphoric acid = 37.9 g

Now, we can calculate the moles:

  • Moles of potassium hydroxide = mass / molar mass = 69.9 g / 39.10 g/mol = 1.79 mol
  • Moles of phosphoric acid = mass / molar mass = 37.9 g / 97.99 g/mol = 0.39 mol

Next, we need to determine the stoichiometric ratio between the reactants and the product. From the balanced chemical equation:

2 KOH(aq) + H3PO4(aq) → K2HPO4(aq) + 2 H2O(l)

We can see that the ratio between potassium hydroxide and potassium phosphate is 2:1. Therefore, for every 2 moles of potassium hydroxide, we can form 1 mole of potassium phosphate.

Comparing the moles of potassium hydroxide and phosphoric acid, we can see that the moles of phosphoric acid (0.39 mol) are less than half of the moles of potassium hydroxide (1.79 mol). This means that phosphoric acid is the limiting reagent.

To calculate the maximum amount of potassium phosphate formed, we use the stoichiometric ratio:

1 mole of potassium hydroxide → 0.5 moles of potassium phosphate

Since phosphoric acid is the limiting reagent, we use its moles to calculate the maximum amount of potassium phosphate:

0.39 mol H3PO4 × (0.5 mol K2HPO4 / 2 mol H3PO4) = 0.0975 mol K2HPO4

Therefore, the maximum amount of potassium phosphate that can be formed is 0.0975 moles.

The formula for the limiting reagent is phosphoric acid (H3PO4).

To calculate the amount of excess reagent remaining, we subtract the moles of the limiting reagent from the moles of the excess reagent:

Moles of excess reagent = Moles of potassium hydroxide - Moles of phosphoric acid

Moles of excess reagent = 1.79 mol - 0.39 mol

Moles of excess reagent = 1.40 mol

Therefore, after the reaction is complete, there will be 1.40 moles of excess potassium hydroxide remaining.

Learn more about calculating the maximum amount of product formed and identifying the limiting reagent here:

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