Asked by Amanda
When 4.5 mol Al reacts with 11.5 mol HCl,what is the limiting reactant, and how many moles of H2 can be formed?
Answers
Answered by
DrBob222
LR = limiting reagent.
2Al + 6HCl ==> 3H2 + 2AlCl3
How many mols H2 produced if we had 4.5 mol Al and all of the HCl we needed. That would be
4.5 x (3 mols H2/2 mol Al) = 4.5 x 3/2 = 6.75.
How many mols H2 produced if we had 11.5 mols HCl and all of the Al we needed. That's
11.5 x (3 mols H2/6) = 11.5 x 3/6 = 4.75 mols H2.
Obviously, both answers can't be right; in LR problems the correct answer is ALWAYS the SMALLER value AND the reagent providing that smaller value is the LR. So you will have 5.75 mols H2 produced.
2Al + 6HCl ==> 3H2 + 2AlCl3
How many mols H2 produced if we had 4.5 mol Al and all of the HCl we needed. That would be
4.5 x (3 mols H2/2 mol Al) = 4.5 x 3/2 = 6.75.
How many mols H2 produced if we had 11.5 mols HCl and all of the Al we needed. That's
11.5 x (3 mols H2/6) = 11.5 x 3/6 = 4.75 mols H2.
Obviously, both answers can't be right; in LR problems the correct answer is ALWAYS the SMALLER value AND the reagent providing that smaller value is the LR. So you will have 5.75 mols H2 produced.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.