Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure. Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.
A) Chlorine is the limiting reactant; 1.14 g of potassium remain.
B) Potassium is the limiting reactant; 2.00 g of chlorine remain.
C) Potassium is the limiting reactant; 2.47 g of chlorine remain.
D) Chlorine is the limiting reactant; 3.07 g of potassium remain.
E) Potassium is the limiting reactant; 4.50 g of chlorine remain.

Question

Quizplus · Accepted Answer

First, we need to write a balanced equation for the reaction:

2 K + Cl2 -> 2 KCl

Next, we need to determine the limiting reactant:

Using the given masses, we can calculate the number of moles of each reactant:

n(K) = m(K) / M(K) = 5.00 g / 39.10 g/mol = 0.128 mol 
n(Cl2) = m(Cl2) / M(Cl2) = 7.00 g / 70.90 g/mol = 0.099 mol

According to the stoichiometry of the balanced equation, 1 mole of Cl2 reacts with 2 moles of K, therefore 0.198 moles of K would be needed to react with 0.099 moles of Cl2. Since we only have 0.128 moles of K, it is the limiting reactant.

Now we can calculate the mass of the excess reactant:

n(Cl2) excess = n(Cl2) - (n(K) x 1/2) = 0.099 mol - (0.128 mol x 1/2) = 0.035 mol 
m(Cl2) excess = n(Cl2) excess x M(Cl2) = 0.035 mol x 70.90 g/mol = 2.47 g

Therefore, the correct answer is C: Potassium is the limiting reactant; 2.47 g of chlorine remain.

Potassium Chloride Is Used as a Substitute for Sodium Chloride