Question

@surry99

Answer 1

ok, first thing I need you to do is calculate the number of moles for all the compounds they gave you weights for...Remember n=m/M ....number of mols = mass/ molar mass

Answer 2

okay.

Answer 3

take your time, you need to get them right ...be back in 5 minutes

Answer 4

I am back.

Answer 5

for the first one, i got approx 2.74
for the second 6.08

Answer 6

wait..i didn't do the third one..ugh

Answer 7

4.38 for the third

Answer 8

ok, to summarize:
2.74 mol of AL2()SO4)3
6.08 mol of Ca(OH)2
4.38 mol of CaSO4 ...correct

Answer 9

yay. next step?

Answer 10

The LR by definition does three things:
1) it is entirely consumed in a reaction
2) it limits or determines the amount of XS react which reacts
3) it limits or determines the amount of product that forms
have you seen these things?

Answer 11

Having sorted out the formulae, balance the equation:

Al2(SO4)3 + 3Ca(OH)2 --> 3CaSO4 + 2Al(OH)3

Now look closely at this equation:
1mol Al2(SO4)3 will react with 3mol Ca(OH)2 to produce 3mol CaSO4 and 2 mol Al(OH)3

: What is the limiting reactant :
Molar mass Al2(SO4)3 = 342.15g/mol
500g = 500/342.15 = 1.461 mol Al2(SO4)3

This will require 3*1.461 = 4.383 mol Ca(OH)2

Molar mass Ca(OH)2 = 74.09 g/mol
450g = 6.074 mol Ca(OH)2 You need 4.383mol to react completely with the Al2(SO4)3, so the Ca(OH)2 is in excess, and the Al2(SO4)3 is the limiting reactant.

Note: there are a number of ways of finding the limiting reactant, but I find that this method is easy and logical.

How many moles of excess reactant are unreacted?
You have 6.074mol. You need 4.383mol : Excess unreacted: 6.074-4.383 = 1.69mol Ca(OH)2 unreacted and remaining.
You have been given the mass of the CaSO4 produced. I guess that this question goes on about % yield, etc. because the mass produced is not required to calculate the limiting reactant.

Answer 12

Oops Formula*

Answer 13

I was typing way past my limit haha ;)

Answer 14

I better check your mole calculations @studygurl14

Answer 15

okay....

Answer 16

Hi @surry99

Answer 17

ok, we need to figure out where your mole calculation went wrong for AL2(SO4)3,,,
for Al2(SO4)3 ...1.46 mol
for Ca(OH)2 ....6.08 mol
Ca(SO4) ....4.38 mol

Answer 18

now again...
The LR by definition does three things:
1) it is entirely consumed in a reaction
2) it limits or determines the amount of XS react which reacts
3) it limits or determines the amount of product that forms
have you seen this before?

Answer 19

yeah, i've seen it befre

Answer 20

don't know where i went wrong...

Answer 21

this is what i did (i rounded)
500/(27(2) + 32.06(3) + 16(3)(4)) = approx 2.74 mol of Al2(SO4)3
450/(40+16(2)+1(2)) = approx 6.08 mol of Ca(OH)2
596/(40+32.06+16(4))= approx 4.38 mol of CaSO4

Answer 22

no problem...ok so by using the mole ratios from the balanced equation I can figure out the LR.

Answer 23

500/(27(2) + 32.06(3) + 16(3)(4)) = approx 2.74 mol of Al2(SO4)3 check the math here later

Answer 24

I am going to assume that the AL2(S04)3 is the LR.
By definition, it would limit or determine the amount of product produced and is entiorely consume in the reaction.
so far so good?

Answer 25

entirely

Answer 26

good

Answer 27

ok so here is how the calculation goes:
1.46 mol of Al2(SO4)3 * 3 mols of CaSO4/1 mol of AL2(SO4)3 = 4.38 mol of CaSO4
Looks like I guessed right...that is the exact amount of product that was produced therefore Al2(SO4)3 is the LR.

Answer 28

You can try later assuming that Ca(OH)2 is the LR and determining how much CaS04 would have been produced but you will find that it would produce more than you actually got , so the assumption that Ca(Oh)2 was the LR must be wrong.

Answer 29

ok..i'm checking my work. still not getting what i did wrong

Answer 30

This method is based on mathematical reasoning ..if a then b...

Answer 31

500/(27(2) + 32.06(3) + 16(3)(4)) = approx 2.74 mol of Al2(SO4)3 check the math here later
when I punch these in I get 1.46 mol

Answer 32

oh, huh. it worked this time

Answer 33

now to figure out how much the Ca(OH)2 is in left over...
Initial amount - what reacts (which is based on the LR) = what is left over
6.08 mol of Ca(OH)2 - 1.46 mol of AL2(SO4)3 * 3 mol of Ca(OH)2/1 mol of Al2(SO4)3 = 1.7 mol of Ca(OH)2

Answer 34

To do these types of problems you must know all three aspects of a LR.

Answer 35

um...wow...

Answer 36

wow good or wow bad?

Answer 37

wow good/bad.

Answer 38

i think i get it.

Answer 39

you going to be ok?

Answer 40

i just have to look over the steps one more time

Answer 41

Yes, thank you So, so much @surry99

Answer 42

Definitely a livesaver. I'llleft you know how the test goes. :)

Answer 43

Please do...
make sure equations are balanced
convert masses to moles (n=m/M)
take mol ratios from the balanced equations for calculations
God Luck!

Answer 44

Good Luck! (too!)

Answer 45

Thanks. Now go feed those hugry puppies!