[Johan]

Alright, more problems. I've been having issues doing things the teachers way, so i've been reasoning things completely different in class. Obviously this could cause some mad problems for me if i get stuck..

How many grams of Al(OH)₃ can be produced from 100.mL of .300M Ba(OH)₂?

3Ba(OH)₂ +2AlCl₃---->2Al(OH)₃+3BaCl₂

I did this so far.. please don't hate .

X mol Ba(OH)₂
------------     = .300M Ba(OH)₂ , now i have .0300 mol Ba..so on
.100 L

I see from the equation that its 3:2 Ba:Al.. so i figure.. .0300 x (3/2)= .045mol of AL

.045mol x 78.0g = 3.51g... now.. that is one of the answers but all of her answers are usually common mistakes that students will make.. so it is very important to understand whats going on. Since no one in class does anything the way i do it, i dunno if i managed to mess up like they may have.. : / so does this appear right?

[enahs]

I see from the equation that its 3:2 Ba:Al.. so i figure.. .0300 x (3/2)= .045mol of AL

Ohh, you were doing so good!

Use the powers of dimensional analysis young padawan.
Search google for dimensional analysis if you are not sure what I am talking about.

0.0300 mol Ba *  2 mol Al = mol of Al!
                        3 mol Ba

Think of it another way.
3/2 > unity (unity = 1). Does that make sense chemically?
That is, you are multiplying by something and increasing its number. Does the number of mols increase? (They can, do not always assume that the ratio has to be less then unity, but look at this case at hand and the question).

[Johan]

ahh that is exactly why i posted, i was trying to reason why if it would be 3/2 or 2/3

yea i somewhat understand Dim Analysis, but the really long ones bother me for some reason. So each problem i have to go through some thought process, instead of plug and chug.

short steps make sense to me though, such as the 3/2 2/3 debate, i just forgot i suppose

I was just about to modify this post to add another problem..

What volume of 6.0M HCL will be needed to react with .350 mol of CaCO₃
CaCO₃ + 2HCL ----> CaCl₂ + H₂O +CO₂

I'm not sure but i think this is where my thought process is broken

.350mol CaCO3
---------        =  6.0M HCL
V L

Obviously that wont work up top, but its 1:2 ratio.. so replace it in terms of HCl

.350mol CaCO3 X 2 mol HCl/1 mole CaCO3 = .700mol HCl

.700 / 6.0 = .1166666 L so my answers go to mL 117mL , and that is one of the answers.. just checking to see if i'm getting the hang of these

[Sev]

What volume of 6.0M HCL will be needed to react with .350 mol of CaCO₃
CaCO₃ + 2HCL ----> CaCl₂ + H₂O +CO₂


.350mol CaCO3 X 2 mol HCl/1 mole CaCO3 = .700mol HCl

.700 / 6.0 = .1166666 L so my answers go to mL 117mL , and that is one of the answers.. just checking to see if i'm getting the hang of these

Looks OK to me.