Question

Given the following unbalanced molecular equation: KAl(SO4)•12H2O (aq) + 2BaCl2 (s) -> KCl (aq) + AlCl3 (aq) + 2BaSO4 (s) (I balanced it already)
a) Balance the reaction and calculate the amount of barium chloride needed to react with 25 mL of a 0.10 M alum solution. and b) What is the percent yield of barium sulfate if 1.02 grams is isolated?

Answer 1

@redbeardd

Answer 2

Ok this is pretty much the same as the last one, but the start up is a little different.

You're given volume and Molariy, to you can relate those with the equation
M(molarity) = n(number of moles)/L(volume)
rearrange to find moles:
n=M•L

Once you have the moles of the product you can work backwards to find out exactly how much of that reactant will be needed to make this amount of product.

This is different from the last question in that you won't need to spend time finding out what the limiting reactant is since you already know how much product has been made, does that make sense?

Anyways, you take the moles of product and convert it to moles of reactant via stoichiometric ratio, then convert that to grams using molar mass to find out exactly how much mass of reactant you get. and THAT is your theoretical yield.

They have given you the actual yield, so just use that same equation from earlier:

actual yield - theoretical yield/theoretical yield x 100

Bam! easy peezy :)

Answer 3

Okay! let me do that, one second

Answer 4

Oh sorry I misread, the second part is a little different. just cut out that last part, once you have the grams of reactant calculated, that's the answer for part A.

Then take that amount (grams of barium chloride) and use that to find out the theoretical yield of barium sulfate.

convert grams of barium chloride to moles using molar mass, then convert that to moles of barium sulfate using stoichiometric ratio, then convert that to grams using molar mass.

That will give you your theoretical yield of barium sulfate, use that with the given actual yield to find percent yield for part B

Answer 5

For the first part, do I just do 0.025L * 0.10 M? or do I need to do something with the M?

Answer 6

thats it just multiply it because the units for M are mol/L, so the L will cancel out and leave you with mol

Answer 7

Sweet. Any idea how I can check my answer for A? It seems kind of large, but I think I did it right

Answer 8

and I don't quite understand part B

Answer 9

Post your math and I'll confirm it for you.

What don't you understand with part B?

You'll be continuing with the answer you got from part A

Answer 10

n = 0.025L * 0.10 M = 0.0025 mol Alum
0.0025 mol x 474.3g Alum/1 mol Alum x 2 mol BaCl2/ 1 mol Alum x 208.2 g BaCl2/ 1 mol BaCl2 = 494 g BaCl2

Answer 11

is 0.0025 mol of Alum the actual yield or theoretical yield?

Answer 12

Redbeardd?

Answer 13

@redbeardd I don't mean to rush you but I have to get up in a few hours and I really want to finish this..

Answer 14

where did you get 474.3g from?

Answer 15

you're using KAl(SO4).12H2O right?

Answer 16

474.3g is the atomic mass of KAl(SO4).12H2O

Answer 17

@dan815 @zepdrix @confluxepic @Luigi0210
Can one of ya'll help, I don't wanna leave my friend hanging, but I'm exhausted and I have to finish a paper.

Answer 18

http://www.lenntech.com/calculators/molecular/molecular-weight-calculator.htm

Answer 19

Oh I'm sorry!

Answer 20

It's not working for me..but I calculated the mass earlier and that's what I got

Answer 21

Sorry for being late. I wasn't online 9 hours ago. @LifeIsADangerousGame

Answer 22

Hey no problem! Thanks for being willing to help. ^_^ I was able to figure it out cx