[alwaysalone]

 5.00 g of an impure mineral which is composted by S in the form of As2S5, were treated with HNO3 until all the S was changed in H2SO4. After this, it was obtained a precipitated in the form of BaSO4 which weight was 0.752 g. What is the % of As2S5 in the impure mineral?
Can you help me? I need the formulas of all the process.
Thanks.

[Borek]

You have to show your attempts at solving the problem to receive help. This is a forum policy.

Hint: you start with a sulfide, you end precipitating BaSO₄. Can you try to guess what is being converted to what?

[alwaysalone]

Thank you for the tip. I had to have seen it by myself. This is my particular solution. Maybe there is someone interested in it.
As₂S₅+40HNO₃ :rarrow:2H₃AsO₄+5H₂SO₄+40NO₂+12H₂O (redox)
H₂SO₄+Ba(OH)₂ :rarrow:BaSO₄+ 2H₂O
0.752gBaSO₄*(1mol BaSO₄/233g BaSO₄)*(1mol S/1 mol BaSO₄)*(32g S/1 molS)=0,10g S
0.10g S*(1mol S/32g S)*(1mol AS₂S₅/5 mol S)*(310 g As₂S₅/1 mol As₂S₅≈=0.2 g As₂S₅
%=0.2+100/5≈4%

[AWK]

Shortcut - counting only S
As₂S₅ 5H₂SO₄ 5BaSO₄

%=0.2+100/5≈4%

Result is OK but with mathematical signs something is wrong.
I am afraid the mineral contains only 4 % of arsenic sulfide contains also much more other sulfides.