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### Topic: Problem of the week - 18/06/2012  (Read 10188 times)

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#### Borek

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##### Problem of the week - 18/06/2012
« on: June 18, 2012, 05:47:30 AM »
Compound A is made of two elements, and at STP is a gas with density 1.232 g/L. One of the products of the reaction of A with water is an acid Ac. Solubility of acid Ac at 30°C is 6.4g/100g of water.

What is mass of crystals of acid Ac produced when 2 moles of compound A react with 500 mL of water?

Edit: wording slightly changed.
« Last Edit: June 18, 2012, 11:24:20 AM by Borek »
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#### AWK

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##### Re: Problem of the week - 18/06/2012
« Reply #1 on: June 18, 2012, 09:36:52 AM »
Quote
- between other products -
If different compounds are possible, the final yield cannot be calculated.
I mean the question concerns the temperature of 30 C and complete reaction with water?
AWK

#### Borek

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##### Re: Problem of the week - 18/06/2012
« Reply #2 on: June 18, 2012, 11:25:32 AM »
I have changed the wording slightly.
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#### Anthasci

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##### Re: Problem of the week - 18/06/2012
« Reply #3 on: June 18, 2012, 07:09:40 PM »
96,6 grams, assuming water density = 1g/mL

#### Schrödinger

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##### Re: Problem of the week - 18/06/2012
« Reply #4 on: June 19, 2012, 02:58:44 AM »
Wild guess : CO(Carbon monoxide) or NO(Nitric oxide)? (One of these?) (Just left CO there, although I felt it doesnt really fit the profile... wasn't sure)
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#### Borek

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##### Re: Problem of the week - 18/06/2012
« Reply #5 on: June 19, 2012, 03:34:41 AM »
No and no.
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#### DrCMS

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##### Re: Problem of the week - 18/06/2012
« Reply #6 on: June 19, 2012, 04:45:28 AM »
I'll try 229.1g as the mass of undissolved solid

#### Rutherford

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##### Re: Problem of the week - 18/06/2012
« Reply #7 on: June 19, 2012, 02:41:46 PM »
Is it 22.98g?

#### Borek

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##### Re: Problem of the week - 18/06/2012
« Reply #8 on: June 20, 2012, 06:50:24 PM »
I'll try 229.1g as the mass of undissolved solid

Compound A is B2H6, acid is H3BO3. Calculation of the mass is a little bit tricky because of two easy to make mistakes. First, not all acid precipitates - some stays dissolved. Second, amount of water that is a solvent is not 500 mL, as it partially reacted.
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#### DrCMS

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##### Re: Problem of the week - 18/06/2012
« Reply #9 on: June 21, 2012, 04:55:34 AM »
Here's how I did it for anyone who's interested.

From the density at STP I calculated the molecular weight 22.4x1.232 = 27.6

Given the info it was a gas comprised of only 2 elements limited the options for the calulated molecular weight to CO, C2H4 or B2H6.

As far as I knew of those three only the diborane would react with water to give an acid.

B2H6 + 6H2O     2H3BO3 + 6H2

That matched the question giving the acid as only one of the products of the reaction with water, hydrogen being the other.

A quick google showed the boric acid solubilty at 25°C was similar to the value given in the problem at 30°C.

Given 2 moles reacting would make 4 moles of boric acid and consume 12 moles of water it was just a case of setting up the equation to calculate the total mass of 4 moles of boric acid and how much water was left for it to dissolve in.  Once I had the amount of boric acid in solution the rest must be the solid formed.

4 moles of boric acid weigh 247.3g
12 moles of water is 216.2g
The water left for the acid to dissolve in is 500-216.2 = 283.8g
Given 6.4g/100ml as the solubilty then 18.2g of the boric acid dissolved so (247.3-18.2) = 229.1g was undissolved solid.

#### Borek

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##### Re: Problem of the week - 18/06/2012
« Reply #10 on: June 21, 2012, 05:09:28 AM »
For the record - its a question from the Polish Chemistry Olympiad.
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#### DrCMS

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##### Re: Problem of the week - 18/06/2012
« Reply #11 on: June 21, 2012, 05:49:28 AM »
It's a nice question; a boron compound was my last thought for a gas of the right weight but once I had that everything else matched up and was reasonably easy.