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). 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

(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