Question

For the equation 2KCI + CaF₂ ⇒ 2 KF + CaCI₂, 55.1 mol KCI is reacted with 46.7 mol CaF₂.

1. What is the limiting reactant?

2. How many moles of KF will be produced?

Please Show all your work!! Thank you!

Answer 1

2

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Finding Limiting reactant:

Convert all reactant data to moles, divide by respective coefficient => smaller value is the limiting reactant.

Given 2KCl + CaF₂ => 2KF + CaCl₂

55.1mole 46.7mole

(55.1/2) = 27.55 (46.7/1) = 46.7

Limiting reactant is KCl b/c smaller value after dividing by respective coefficients. CaF₂ will be in excess when rxn consumes all KCl.

Moles KF = 55.1 mole as coefficients of two substances in balanced equation are equal.