Question

For the following reaction, 53.7 grams of potassium hydroxide are allowed to react with 29.4 grams of phosphoric acid.

potassium hydroxide (aq) + phosphoric acid (aq) potassium phosphate (aq) + water (l)

What is the maximum amount of potassium phosphate that can be formed?  grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?  grams

Answer 1

Balanced chemical reaction;

3KOH + H3PO4 -----> K3PO4 + 3H2O

Moles of KOH = mass/molar mass = 53.7/56.1 = 0.95722

Moles H3PO4 = mass/molar mass = 29.4/98 = 0.3 moles

From reaction;

1 mole H3PO4 requires 3 mole KOH

So, 0.3 mole H3PO4 will require = 3*0.3 = 0.9 moles of KOH

However, available moles of KOH = 0.95722 which is greater than required moles i.e. 0.9

So, unreacted moles of KOH = 0.95722 - 0.9 = 0.05722 moles

So, mass or amount of excess reagent i.e. KOH remains after the reaction = remaining moles * molar mass = 0.05722*56.1 = 3.21 grams ......Answer

Thus, KOH is excess reagent and H3PO4 is limiting reagent. Limiting reagent will drive the yield of product.

Again from reaction;

1 mole H3PO4 produces 1 mole of K3PO4

So, 0.3 mole H3PO4 will produce = 1*0.3 = 0.3 moles of K3PO4

Maximum amount formed of K3PO4 = moles * molar mass = 0.3 * 212.266 = 63.68 grams .....Answer