[zpmodel]

I have a problem and can not solve it:

A piece of metallic aluminum weighing 2.50 grams is treated with 75.0 mL of sulphuric acid ( d = 1.18 g/cm3 , 24.7% H₂SO₄ ). After the metal is dissolved, the solution is diluted to 400mL.

a) Calculate the morality of the solution with respect to the aluminum salt formed.
b) Calculate the morality of the resulting solution with respect to unreacted sulphuric acid.


I made a slight attempt, but got lost after it:

75ml x 1.18g = 88.5g

Xg / 88.5g = .247
Xg = 21.859g of H₂SO₄

I actually don't even know what I am doing anymore
I'm seeking a learning experience so that I can solve other problems as well. Please, any help would be greatly appreciated.

[nj_bartel]

?Al + ?'H₂SO₄ -> .?

(g Al) / (MM Al) = ? moles Al       (mL H₂SO₄)(D H₂SO₄) / (MM H₂SO₄) = ? moles H₂SO₄


Work with that and see where it takes you.

[Borek]

Xg / 88.5g = .247
Xg = 21.859g of H₂SO₄

That's correct mass of the sulfuric acid used.

This is a limiting reagent question.

[zpmodel]

You know, I think the answer that I was given along with the question is wrong. At the back of my book it says that the answer is:

0.21 mol/L H₂SO₄
0.116 mol/L Al₂(SO₄)₃

For the morality of the solution with respect to the aluminum salt formed, it would be the number of moles of the limiting reagent, Al, which is 0.09, divided by the number of litres, which is .4

And for the morality of the resulting solution with respect to unreacted sulphuric acid, it would be the difference in moles that was used for the aluminum salt formed, which would be 0.22 - 0.09. So the morality here would be .13 divided by the number of liters, which is .4

That's the most sense I could make out of it. But I think I should have a balanced equation first:

2Al + 6H₂SO₄   -->  2Al₂(SO₄)₃ + 6H₂

Is this correct for the equation?

[Borek]

Check Al atoms in the equation.

Your final concentration estimation is interesting, but wrong - how many moles of Al per mole of aluminum sulfate?

[zpmodel]

Check Al atoms in the equation.

Your final concentration estimation is interesting, but wrong - how many moles of Al per mole of aluminum sulfate?

Oh yeah, It's not balanced :p
Here we go:

2Al + 3H₂SO₄   -->  Al₂(SO₄)₃ + 3H₂

It's a ratio of 2 moles of Al to 1 mole of aluminum sulfate.
So the number of moles of aluminum sulfate would be:

(0.09 / 2) = 0.045 moles

So the concentration would be

0.045 / .4 = 0.1125 m/L

Which is pretty close to the answer, but not exact...

As for the morality of the resulting solution with respect to unreacted sulphuric acid, I would subtract the used moles of Al from the total amount of moles of sulphuric acid, right? So I do:

.22 mol / 3 = .073 moles of sulphuric acid.
.073 mol - 0.045 mol = .0283 moles of sulphuric acid left over after reaction.

.0283 mol / .4L = .071 m/L

I must be doing something terribly wrong...

[enahs]

With respect A; Only round at the end, not in intermediate steps. You are compounding your rounding errors.

B:
1)How many mols of Al?
2) How many mols of sulfuric acid?
3)Which is the limiting reagent?
4)If all of the limiting reagent is used up, how much of the other reagent is required for the reaction?
5) How much is left of the excess reagent? (step 2 - step 4).

[zpmodel]

B:
1)How many mols of Al?
2) How many mols of sulfuric acid?
3)Which is the limiting reagent?
4)If all of the limiting reagent is used up, how much of the other reagent is required for the reaction?
5) How much is left of the excess reagent? (step 2 - step 4).

1) .09 / 2 = .05 moles of Al
2) .22 / 3 = .07 moles of sulfuric acid.
3) Al is the limiting reagent.
4) I don't understand.
5) .07 - .05 = .02 moles of sulfuric acid

[enahs]

No.
1)
2.5g Al / (26.982 g/mol) = 0.09265 mol Al. That is it.
2)
21.859 g H2SO4 / (98.078 g/ mol) = 0.222 mols H2SO4. That is it.
3)
Correct. Why?
4)
According to your balanced equation, for every 2 mols of aluminum used, 3 mols of H2SO4 is required.
0.09265 mol Al *    3 mol H2SO4    =  0.1389 mol H2SO4
                            2 mol Al

5)
If you started with 0.222 mols H2SO4, and used 0.1389, you are left with 0.222-0.1389 = 0.0830 mols H2SO4 left.
If that is in 400 mL, the molarity is mols/L = 0.083mols/0.4 L = 0.21M

[zpmodel]

According to your balanced equation, for every 2 mols of aluminum used, 3 mols of H2SO4 is required.
0.09265 mol Al *    3 mol H2SO4    =  0.1389 mol H2SO4
                            2 mol Al

THANK YOU!!!
That actually makes a lot of sense, and clears things up