Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on June 18, 2012, 05:47:30 AM
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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.
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- 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?
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I have changed the wording slightly.
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96,6 grams, assuming water density = 1g/mL
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Wild guess : CO(Carbon monoxide) or NO(Nitric oxide)? (One of these?) :P (Just left CO there, although I felt it doesnt really fit the profile... wasn't sure)
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No and no.
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I'll try 229.1g as the mass of undissolved solid
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Is it 22.98g?
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I'll try 229.1g as the mass of undissolved solid
And that's the correct answer :)
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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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 :rarrow: 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.
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For the record - its a question from the Polish Chemistry Olympiad.
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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.