Question

Limiting and Excess Reactant

1. Aluminum chloride, AlCl3, can be made by the reaction of aluminum with chlorine according to the following equation (must be balanced first):
2Al + 3Cl2 ------> 2AlCl3
What are the limiting and excess reactants if 20.0 grams of Al and 30.0 grams of Cl2 are used?

2. Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 25 grams of chloroform and 0.36 mol of chlorine were mixed. Which is the limiting reactant? The equation is given as
CHCl3 + Cl2 -------> CCl4 + HCl

Answer 1

1. LR: Cl2, ER: Al 2. LR: CHCl3

Step-by-step explanation:

1. Find g/moles of Al and Cl2

g/moles of Al: 0.74 and Cl2 is 0.42

We can find the limiting reactant by take those number divided by 2 and 3

0.74 divided by 2 = 0.37

0.42 divided by 3 = 0.14 so 0.14 is the least -> that going to be the limiting reactant. (Cl2)

Now we know Al is a excess reactant.

2. That equation is already balanced and it's already give us 0.36 mole of chlorine, now we just need to find the mole of chloroform ( CHCl3 )

To do that we need to find the molar mass of CHCl3 which is 119.0 g/moles

then take 25 grams ÷ by 119g/moles we will get: 0.21 mole of CHCl3 that's mean CHCl3 is a limiting reactant.