Question

Balance the following equation,

bH3PO4 + aKOH --> cK3PO4 + dH2O
And answer the following questions:
Given 5.0 moles of KOH and 2.0 moles of H3PO4, how many moles of K3PO4 can be prepared? 1.27moles of K3PO4 can be formed.
In the above problem, which reactant is the limiting reagent? Which is the excess?

SHOW ALL WORK AND STEPS - VIA WORDS OR CALCULATIONs

Answer 1

a. H3PO4 + 3KOH => K3PO4 + 3H2O

b. 1.67 moles

c. KOH is the limiting agent

Step-by-step explanation:

H3PO4 + 3KOH => K3PO4 + 3H2O

a = 3, b = 1, c = 1, d = 3

1 mole of H3PO4 will react with 3 moles of KOH

=> 2 moles of H3PO4 will react with 6 moles of KOH

since there are only 5 moles of KOH available, KOH is the limiting agent

Since there are only 5 moles of KOH available, the amount of H3PO4 needed is (5 x 1)/3 = 5/3 = 1.67 moles

1 mole of H3PO4 will produce 1 mole of K3PO4 => 1.67 moles of H3PO4 will produce 1.67 moles of K3PO4

Hope this helps