Potassium hydroxide and phosphoric acid react to form potassium phosphate and water according to the equation:

3KOH (aq) + H3PO4 (aq) K3PO4 (aq) + 3H2O (l)

Determine the starting mass of each reactant if 62.1 g of K3PO4 is produced and 80.8 g of H3PO4 remains un reacted.

A) what is the starting mass of KOH?

B) What is the starting mass of H3PO4?

(a) Starting mass of KOH is 187.7g

(b) Starting mass of H3PO4 is 109.5g

Explanation:

From the equation of reaction,

1 mole (98g) of H3PO4 was used up to produce 1 mole (212g) of K3PO4

Mass of H3PO4 used up to produce 62.1g of K3PO4 = 98*62.1/212 = 28.7g of H3PO4

Starting mass of H3PO4 = mass of H3PO4 used up + mass of unreacted H3PO4 = 28.7g + 80.8g = 109.5g

Also, from the equation of reaction,

3 moles (168g) of KOH reacted with 1 mole (98g) of H3PO4

Starting mass of KOH that reacted with 109.5g of H3PO4 = 168*109.5/98 = 187.7g of KOH

A). The starting mass of KOH = 49.2g

B). The starting mass of H3PO4 = 109.5g

Explanation:

Given;

Molar mass of KOH = 56g/mol

Molar mass of H3PO4 = 98g/mol

Molar mass of K3PO4 = 212g/mol

Mass of K3PO4 produced = 62.1g

Mass of unreacted H3PO4 = 80.8g

Mole ratio of KOH : H3PO4 : K3PO4 = 3:1:1

Number of moles = mass/molar mass ... 1

Using equation 1.

Number of moles of K3PO4 produced

N = 62.1g/212g/mol = 0.293 mol

Therefore: using the mole ratio in the reaction;

Mass of H3PO4 reacted = number of moles * molar mass ... 2

M = 1*0.293mol*98g/mol

M = 28.7g

Total mass of H3PO4 = mass of reacted + mass of unreacted H3PO4.

Total mass = 28.7g + 80.8g = 109.5g

Starting mass of H3PO4 = 109.5g

Also for KOH,

Mass of reacted KOH = number of moles * molar mass

Since 3 moles of KOH will produce 1 mole of K3PO4.

3 * 0.293 moles of KOH will produce 0.293 moles of K3PO4.

Number of moles of KOH = 3 * 0.293moles = 0.879mole

Mass of reacted KOH = 0.879 mol * 56g/mol = 49.2g

Since there are no unreacted KOH.

Starting mass of KOH = 49.2g