Question

When 3.5 mol Al react with 12.5 mol HCl, what is the limiting reactant and how many moles of H2 can be formed?

2 Al + 6 HCl “yields”/ 2 AlCl3 + 3 H2

Answer 1

i only have math problems left now .-. im no good at math

Answer 2

2Al require 6HCl

that means, 1 Al require 3 HCl, right ?

Answer 3

yus

Answer 4

you got 3.5 mol Al, so lets see if we have sufficient HCl for this :

1 Al require 3 HCl
3.5Al require 3.5*3 HCl

Answer 5

so itd require 7.0 HCl

Answer 6

whats the value of 3.5*3 ?

Answer 7

so al would be the limiting reactant since we dont have enough of it

Answer 8

whats the value of 3.5*3 ?

Answer 9

i already said XD itd be 7.0 HCl

Answer 10

check again :o

Answer 11

3.5*3 = 3.5 + 3.5 + 3.5
= ?

Answer 12

oh itd be 10.5

Answer 13

yes :)
but still you're right about the limiting reactant :

Al is the limiting reactant because it will get used up first leaving 12.5-10.5 = 2mol of HCl unused

Answer 14

so then 3.5 mols of H2 owuld be formed

Answer 15

nope

Answer 16

3.5mol of Al is getting reacted,
lets find out how many mols of H2 will be foremed

Answer 17

given equation :

`2 Al` + 6 HCl “yields”/ 2 AlCl3 + `3 H2`

Answer 18

Notice that,
2Al is producing 3H2

that means,

1Al will produce 3/2 H2, right ?

Answer 19

yes...

Answer 20

that also means,

3.5 Al will produce 3.5*(3/2) H2, yes ?

Answer 21

yes, thad mean that 5 1/4 would be produced

Answer 22

Correct !