Answer to Question #155513 in General Chemistry for Sage Topaz

Question

Gold resists most attacks by most reactants. But hot Chlorine gas is chemically active enough to react with it. At 150 degree celsius the following reaction takes place:

                       2 Au + 3 Cl₂ → 2 AuCl₃

Suppose 10 g of gold and 10 g of Cl₂ are sealed in container and heated until the reaction is complete, what is the limiting reactant? How much of the product is formed? How much excess or unreacted substance is left?

Answer 1

The limiting reactant is a reactant that is fully consumed when the chemical reaction is completed. To determine the limiting reactant, we need to calculate the amounts of each reactant: 

Molar mass of gold: MW(Au) = 197.0 g/mol 

Amount of the gold: n(Au) = m(Au)/M(Au) = 10 g / 197.0 g/mol = 0.051 mol

Molar mass of the chlorine gas: MW(Cl₂) = 70.9 g/mol 

Amount of the chlorine gas: n(Cl₂) = m(Cl₂)/M(Cl₂) = 10 g / 70.9 g/mol = 0.141 mol

According to the reaction equation, for a complete reaction of 2 moles of gold, we need 3 moles of chlorine gas, or, that is the same, 1 mole of gold for 1.5 moles of chlorine gas. According to this ratio, for 0.051 mol of gold, 0.051*1.5=0.077 mol of chlorine gas was consumed, and 0.141-0.077=0.064 mol was left unreacted. Since the amount of chlorine gas that we have is larger than was consumed, the gold is the limiting reactant. The mass of the product calculation:

Molar mass of gold chloride: MW(AuCl3) = 303.3 g/mol 

According to the equation the amount of gold chloride is the same as gold: n(AuCl3) = n(Au) = 0.051 mol

The mass of the gold chloride obtained: m(AuCl₃) = n(AuCl₃)*M(AuCl₃) = 0.051 mol * 303.3 g/mol = 15.47 g

The mass of unreacted chlorine gas can be calculated from the amount of unreacted chlorine gas:

m(Cl₂) = n(Cl₂)*M(Cl₂) = 0.064 mol * 70.9 g/mol = 4.538 g