10g of alumium reats with 35g of chlorine gas to produce alumium chloride. Which reactant is limiting, which is in excess, and how much product is produced? (Al:27g/mol, AlCl3:166.5g/mol)
2Al + 3Cl2 -> 2AlCl3
2 answers
alcl3:166.5g/mol (X) -> 133.5g/mol
2Al + 3Cl2 -> 2AlCl3
Here is how you work these limiting reagent problems (LR) but I must be honest and say I do them the long way.
I've estimated all calculations so you will need to recalculate all of them.
Step 1. Convert what you have to grams.
a. mols Al = 10/27 = about 0.4
b. mols Cl2 = 35/35.5 = about 0.5
Step 2. Convert mols Al and mols Cl2 to mols AlCl3 using the coefficients in the balanced equation.
a. 0.4 mols Al x (2 mols AlCl3/1 mols Al) = about 0.4
b. 0.5 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = about 0.33
c. In LR problems, the small number always wins. Do you understand why? So Cl2 is the LR and Al is the ER (excess reagent)/
Step 3. mols AlCl3 x molar mass AlCl3 = grams AlCl3.
Here is how you work these limiting reagent problems (LR) but I must be honest and say I do them the long way.
I've estimated all calculations so you will need to recalculate all of them.
Step 1. Convert what you have to grams.
a. mols Al = 10/27 = about 0.4
b. mols Cl2 = 35/35.5 = about 0.5
Step 2. Convert mols Al and mols Cl2 to mols AlCl3 using the coefficients in the balanced equation.
a. 0.4 mols Al x (2 mols AlCl3/1 mols Al) = about 0.4
b. 0.5 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = about 0.33
c. In LR problems, the small number always wins. Do you understand why? So Cl2 is the LR and Al is the ER (excess reagent)/
Step 3. mols AlCl3 x molar mass AlCl3 = grams AlCl3.
2 answers