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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: alwaysalone on June 20, 2019, 01:28:29 PM

Title: Molar relations
Post by: alwaysalone on June 20, 2019, 01:28:29 PM

 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.

Title: Re: Molar relations
Post by: Borek on June 20, 2019, 03:55:12 PM

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?

Title: Re: Molar relations
Post by: alwaysalone on June 22, 2019, 06:36:25 AM

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%

Title: Re: Molar relations
Post by: AWK on June 22, 2019, 06:49:19 AM

Shortcut - counting only S
As₂S₅ :rarrow: 5H₂SO₄ :rarrow: 5BaSO₄

Quote

%=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.