October 15, 2019, 08:19:02 PM
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### Topic: Molar relations  (Read 389 times)

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#### alwaysalone

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##### Molar relations
« 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.

#### Borek

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##### Re: Molar relations
« Reply #1 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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### alwaysalone

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##### Re: Molar relations
« Reply #2 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%

#### AWK

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##### Re: Molar relations
« Reply #3 on: June 22, 2019, 06:49:19 AM »
Shortcut - counting only S
As2S5 5H2SO4 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.
AWK