[Evaldas]

Hi again! I would like to ask if you know a simpler way to solve this exercise:
Into 500 ml of 0.2 mol/l concentration barium chloride 88.5 mililitres of 14.2% sodium sulphate (ϱ=1.13 g/cm³) had been added. Count how many grams of percipitates had formed (in grams).

My long solution (with some comments):
So first the reaction:
2BaCl₂(aq) + 2Na₂SO₄(aq)  2BaSO₄(s) + 4NaCl(aq)
So we have:
V_(BaCl2) = 500 ml
c_(BaCl2) = 0.2 mol/l

V_(Na2SO4) = 88.5 ml
c_(Na2SO4) = 14.2%
ϱ_(Na2SO4) = 1.13 g/cm³

And some other things that are practical to count:
M_(BaCl2) = 208 g/mol
M_(BaCl2) = 142 g/mol
M_(BaSO4) = 233 g/mol

We need to find:
m_(BaSO4) - ?

Solution:
1) The mass of BaCl₂ in grams:
c = n/V  => n = c x V = 0.2 mol/l x 0.5 l = 0.1 mol => m = n x M = 0.1 mol x 208 g/mol = 20.8 g
2) The mass of Na₂SO₄ in grams:
(so first volume of plain sodium sulphate)
V = 88.5 x 0.142 = 12.567 ml => m = 1.13 g/cm³ x 12.567 ml = 14.2 g
n = 14.2 g / 142 g/mol = 0.1 mol
3) n_(BaSO4) = n_(BaCl2) = n_(Na2SO4) = 0.1 mol

m_(BaSO4) = 0.1 mol x 233 g/mol = 23.3 g

Here. So my answer matches with the given, but I think my solution is too long! Please help...

[Borek]

Your reaction equation - while balanced - is technically incorrect, as these are not the smallest possible coefficients. It doesn't change the final result, howoever.

1) The mass of BaCl₂ in grams:
c = n/V  => n = c x V = 0.2 mol/l x 0.5 l = 0.1 mol => m = n x M = 0.1 mol x 208 g/mol = 20.8 g

Do you need it for anything?

2) The mass of Na₂SO₄ in grams:
(so first volume of plain sodium sulphate)
V = 88.5 x 0.142 = 12.567 ml => m = 1.13 g/cm³ x 12.567 ml = 14.2 g

This acidentally gives correct result, but is wrong. You should calculate mass of the solution first, mass of the sulfate next. 12.567 mL doesn't make sense - mL of what?

[Evaldas]

First, yes, my bad:
BaCl₂(aq) + Na₂SO₄(aq)    BaSO₄(s) + 2NaCl(aq)

Second, you're right - no I do not need mass of BaCl₂

It's 12.567 of plain Na₂SO₄ in the 88.5 ml of 14.2% concentration solution:
V = 88.5 ml* x 0.142** = 12.576 ml
* - the volume of the aqueos solution
** - 0.142 because we have 14.2% of Na₂SO₄ in that aqueos solution. 1% of 88.5 would be 0.885 ml, so 0.885 x 14.2% = 12.576 ml.
Or am I speaking non-sense ?

[Borek]

Yes, you are speaking nonsense 

According to wiki density of pure sodium sulfate ranges between 2.664 g/cm3 (anhydrous) and 1.464 g/cm3 (decahydrate) - I am not 100% sure of these numbers, but they sound reasonable.

14.2% solution usually means that 14.2 % of the solution mass is solute. What you should do is to calculate mass of the solution - 88.5 mL * 1.13 g/mL = 100 g; then mass of the sodium sulfate - 100g * 14.2% = 14.2 g.

[Evaldas]

But this is the density they give in the workbook.

Ah, I think I'm getting it now, I thought it was the density of pure sodium sulfate, but I realised that they meant it was the density of 14.2% sodium sulfate solution. But now the answer would change and wouldn't match the one given...

But anyways, speaking of simpler ways, I meant one where they write moles underneath the reaction according to the coefficients, something like this:
BaCl₂(aq) + Na₂SO₄(aq)  BaSO₄(s) + 2NaCl(aq)
   1 mol         1 mol             1 mol          2 mol
And then they write above what masses they have, and write an 'x' above of what they're trying to find, and then they solve it with a proportion...
Do you know what I'm talking about?

[Borek]

Ah, I think I'm getting it now, I thought it was the density of pure sodium sulfate, but I realised that they meant it was the density of 14.2% sodium sulfate solution. But now the answer would change and wouldn't match the one given...

No, accidentally it will give the same answer.

But anyways, speaking of simpler ways, I meant one where they write moles underneath the reaction according to the coefficients, something like this:
BaCl₂(aq) + Na₂SO₄(aq)  BaSO₄(s) + 2NaCl(aq)
   1 mol         1 mol             1 mol          2 mol
And then they write above what masses they have, and write an 'x' above of what they're trying to find, and then they solve it with a proportion...
Do you know what I'm talking about?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions