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.

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

2Al(s)+3Cl2(g)→2AlCl3(s)

You are given 34.0 g of aluminum and 39.0 g of chlorine gas.

PART A QUESTION:

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 34.0 g of aluminum?

Express your answer to three significant figures and include the appropriate units.

PART B QUESTION:

Part B

If you had excess aluminum, how many moles of aluminum chloride could be produced from 39.0 g of chlorine gas, Cl2?

Express your answer to three significant figures and include the appropriate units.

Answer 1