Question

When potassium hydroxide reacts with phosphoric acid, potassium phosphate and water form. Suppose 6.26 mol of potassium hydroxide are reacted with 2.36 moo of phosphoric acid.

Determine the limiting reagent and the reagent in excess.

Answer 1

Answer on the question #68144, Chemistry / Other

Question:

When potassium hydroxide reacts with phosphoric acid, potassium phosphate and water form. Suppose 6.26 mol of potassium hydroxide are reacted with 2.36 moo of phosphoric acid.

Determine the limiting reagent and the reagent in excess.

Answer:

Let's write the equation of reaction:

$$3KOH + H_3PO_4 \rightarrow K_3PO_4 + 3H_2O$$

As one can see, 3 moles of potassium hydrohyde react with 1 mole of phosphoric acid. Thus, the relation between the number of the moles of $KOH$ and $H_3PO_4$ should be:

$$\frac{n(KOH)}{3} = n(H_3PO_4)$$ $$\frac{n(KOH)}{3} = \frac{6.26 \text{ mol}}{3} = 2.087 \text{ mol}$$ $$n(H_3PO_4) = 2.36 \text{ mol}$$

2.087 > 2.36, so phosphoric acid is in excess.

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