# Problem: 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 27.0 g of aluminum with 32.0 g of chlorine?In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything leftover:2A + B → A2BBut 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 A2B3.2 mol B 1 mol A2B1 mol B= 3.2 mol A2BNotice 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 amountsAluminum reacts with chlorine gas to form aluminum chloride via the following reaction:2 Al(s) + 3 Cl2 (g) → 2 AlCl3(s)You are given 27.0 g of aluminum and 32.0 g of chlorine gas.

###### FREE Expert Solution
90% (76 ratings) View Complete Written Solution
###### Problem Details

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 27.0 g of aluminum with 32.0 g of chlorine?

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

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:

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

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

2 Al(s) + 3 Cl(g) → 2 AlCl3(s)

You are given 27.0 g of aluminum and 32.0 g of chlorine gas.