Question

Limiting Reactant Procedure

In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over:

2A+B→A2B

But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately:

2.8 mol A×1 mol A2B2 mol A=1.4 mol A2B

3.2 mol B×1 mol A2B1 mol B=3.2 mol A2B

Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of A2B can be formed from the given amounts.

PART C:Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

What is the maximum mass of aluminum chloride that can be formed when reacting 34.0 g of aluminum with 39.0 g of chlorine?

Answer 1

calculate mass of AlCl3 formed by Al:

mass of Al = 34 g

molar mass of Al = 26.98 g/mol

mol of Al = (mass)/(molar mass)

= 34/26.98

= 1.26 mol

According to balanced equation

mol of AlCl3 formed = (2/2)* moles of Al

= (2/2)*1.26

= 1.26 mol

mass of AlCl3 = number of mol * molar mass

= 1.26*133.33

= 168 g

calculate mass of AlCl3 formed by Cl2:

mass of Cl2 = 39 g

molar mass of Cl2 = 70.9 g/mol

mol of Cl2 = (mass)/(molar mass)

= 39/70.9

= 0.550 mol

According to balanced equation

mol of AlCl3 formed = (2/3)* moles of Cl2

= (2/3)*0.550

= 0.367 mol

mass of AlCl3 = number of mol * molar mass

= 0.367*133.33

= 48.9 g

mass is less formed in case of Cl2

Hence Cl2 is limiting reagent and mass formed = 48.9 g

Answer: 48.9 g