PROBLEM 47P

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

2 K(s) + Cl2(g) → 2 KCl(s)

(a) 1 mol K; 1 mol Cl2

(b) 1.8 mol K; 1 mol Cl2

(c) 2.2 mol K; 1 mol Cl2

(d) 14.6 mol K; 7.8 mol Cl2

Solution 47P

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

2 K(s) + Cl2(g) → 2 KCl(s)

(a) 1 mol K; 1 mol Cl2

(b) 1.8 mol K; 1 mol Cl2

(c) 2.2 mol K; 1 mol Cl2

(d) 14.6 mol K; 7.8 mol Cl2

The equation tells you that 2 mol K react with 1 mol Cl2

Ratio = 2:1

The theoretical yield is the proportion which gives the least amount of moles of product, and is called as limiting reactant,

a).1 mol K; 1 mol Cl2 : Ratio = 1:1 K is limiting

b).1.8 mol K; 1 mol Cl2 - Ratio 1.8:1 Cl2 is limiting

c).2.2 mol K; 1 mol Cl2 = Ratio 2.2:1 = Cl2 is limiting

d).14.6 mol K, 7.8 mol Cl2 Ratio 14.6:7.8 = 14.6/7.8 = 1.87:1 Cl2 is limiting