Question

When 5.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of AlCl3 can be formed? 2 Al + 6HCl → 2 AlCl3 + 3 H2

A. Al is the limiting reactant, 5.0 mol AlCl3 can be formed

B. HCl is the limiting reactant, 4.3 mol AlCl3 can be formed

C. Al is the limiting reactant, 7.5 mol AlCl3 can be formed

D. HCl is the limiting reactant, 6.5 mol AlCl3 can be formed

Answer 1

please help<3 I'd really appreciate it

Answer 2

Concentrate on the left hand side of the equation, and observe the ratio of reactants.

Answer 3

Sorry I dont understand

Answer 4

According to the equation of the reaction, what is the ratio of Al to HCl?

Answer 5

umm 2:3 ?

Answer 6

How did you arrive at 2:3?

Answer 7

because the right hand side says its 2 AlCl and 3H2

Answer 8

oh wait nvm
it's 2:6

Answer 9

The right hand side refer to products.
We are dealing with reactants, meaning the substances on the left hand side, Al and HCl.
Can you try again?

Answer 10

Yea so it would be 2:6 or 1:4

Answer 11

Yes, 2:6, can you reduce it?

Answer 12

Sorry, \(2:6 \ne 1;4\)! :(

Answer 13

oh lol I mean 1:3

Answer 14

ok!

Answer 15

ok so now what

Answer 16

Now look at the quantity of actual reactants.
Which of Al or HCl will be used up first if the reaction requires a ratio of 1:3?

Answer 17

3 h2?

Answer 18

I think I mean Im not sure

Answer 19

Don't understand what you wrote! :(

Answer 20

Like in the right side, theres 3 H2

Answer 21

So that would be the one that'd be used up first. Idk I'm just guessing

Answer 22

Yes, but do not guess, work it out!

Answer 23

wait so I'm right?

Answer 24

that isn't an option though

Answer 25

Right now, we are not onto the right side yet. We are trying to find the limiting reagent.

Answer 26

oh..ok so then I guess it would be HCl

Answer 27

Limiting reagent is, the one which will be used up first.

Answer 28

ohh nvm so it would be Al

Answer 29

Please, do not guess.
Every answer has a reason behind it.
You cannot guess your way to your degree, there are exams.
If you do not understand, you will have problems later on.

Answer 30

No I know, I'm not guessing. I really think that it's Al

Answer 31

Work it out this way.
The ratio of the equation gives you the ratio of number of moles that will be used up.
1 mole of Al will require 3 moles of HCl.
So how many moles of Al will use up 13 moles of HCl? Use proportions.
1 : 3 = x : 13
what is x.

Answer 32

pellet man. I have no idea how to work that out. I mean, I guess it would be 1 as well? Or 10?

Answer 33

I mean, it might be 10.

Answer 34

Frankly, do you just want the answer, or do you want to know how to solve this problem?

Answer 35

Both would be great to be honest. I just want to finish this test and get on with it. But I don't want to make it seem like I'm lazy or anything. I just need to finish it as soon as possible.

Answer 36

I am sorry, but I don't see efforts in finding out how to solve the problem.
Perhaps someone else will give you the answer, perhaps not.
I wish you luck with your studies.

Answer 37

Please help guys. Ive taken this test 4 times, I still don't understand this question, and I just need someone to assist me.

Answer 38

\[\large \bf 2Al+6HCl \rightarrow 2AlCl_3+3H_2\]

Answer 39

do you know how to find limiting reactant? @brriiarr

Answer 40

no

Answer 41

hello?

Answer 42

so here it is :-
to find limiting reactant,
first find stoichiometry coefficients and then mole of that particular compound and divide mole by stoichiometry coefficient and find the minimum ratio of the all the compounds and compare them and if any there is minimum ratio against all the compounds ,that compound is a limiting reactant.

Answer 43

For better understanding,take example of your question

Answer 44

ahh ok. I sort of remember now.

Answer 45

ready to solve this one ? @brriiarr
and if you have any problem in my step please comment.
ok?

Answer 46

ok yes definitely thank you so much

Answer 47

\[\large \bf \frac{mole}{stoichiometry~coefficient}\]
\[\large \bf Al=\frac{5}{2}=2.5\]
\[\large \bf HCl=\frac{13}{6}=2.16\]

Answer 48

which one is smaller?

Answer 49

HCl

Answer 50

So HCl would be one of the answers, right? Like out of the two HCl options, one is going to be correct?

Answer 51

correct.
So HCl is limiting reactant

Answer 52

ok. now what?

Answer 53

and to find how many moles of compound formed,we use limiting reactant and then balance it.

Answer 54

now i will explain it from your question

Answer 55

ok

Answer 56

hello?

Answer 57

Limiting reactant is HCl,now use this :-
\[\large \bf 6~moles~of~HCl \rightarrow 2~moles~of~AlCl_3\]
\[\large \bf 1~mole~ HCl \rightarrow \frac{2}{6}~moles~AlCl_3=\frac{1}{3}~moles~AlCl_3\]
but we have 13 moles of HCl,
\[\large \bf 13~moles~of~HCl \rightarrow 13 \times \frac{1}{3}~moles~of~AlCl_3\]
so our final answer,
\[\large \bf 13~moles~of~HCl \rightarrow 4.3~moles~of~AlCl_3\]

therefore, 4.3 moles of AlCl3 formed.

Answer 58

From this, option B is correct. :)

Answer 59

ok thank you<3

Answer 60

Welcome :) <3 <3