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Topic: Problem of the week - 30/07/2012  (Read 15591 times)

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Offline Borek

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Problem of the week - 30/07/2012
« on: July 30, 2012, 01:01:00 PM »
When a 1.22 L reaction vessel made of glass was filled with a dry, corrosive gas, its weight increased by 10.2 13.6 g (edit: data corrected) (temperature in the lab was 21 °C). After adding a single drop of water reactor was closed with a manometer. After several hours pressure inside the reactor increased by exactly 50% and the inside surface of glass became matted, with a slight yellow tint. Yellow tint disappeared after the reactor was opened and flushed with a concentrated solution of NaOH.

Name the gas.

Note: I am in a vacation mode, preparing questions in memory and drawing equations on the beach sand*. Chances that I am mistaken are even higher than in a typical POTW.

*An utter lie, no sand here, only rocks. It makes the preparations even more difficult.
« Last Edit: August 14, 2012, 04:18:35 AM by Borek »
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #1 on: July 30, 2012, 07:45:30 PM »
so far:
corrosive gas (A) + water = acid?? (yellow) + gas B

My initial thought is that the corrosive gas maybe a Nitrogen gas.

with the data available, and assuming that the initial pressure (P1) is caused by the gas A only, and the final pressure (P2) is caused by the gas B (product) only:

I have that: P2 = P1X3/2. Since Temp. an volume are constant=> applying the ideal gas law I got that:

Moles B/moles A = 3/2. 

I didn't go further since this maybe completely wrong --feedback???  ???

Offline Borek

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Re: Problem of the week - 30/07/2012
« Reply #2 on: August 01, 2012, 01:21:15 PM »
Assume glass is mainly SiO2.

Nitrogen is not corrosive, it is almost completely inert.
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #3 on: August 01, 2012, 03:12:21 PM »
Assume glass is mainly SiO2.

Nitrogen is not corrosive, it is almost completely inert.

sorry I meant a gas containing N element (an oxide like NX0Y)

Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #4 on: August 01, 2012, 05:10:04 PM »
is it HF?

since it reacts with silica when dissolved in water to form fluorosilicic acid (colorless to light yellow)

SiO2(s) + 6 HF(aq) → H2SiF6(aq) + 2 H2O(l)

Offline Borek

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Re: Problem of the week - 30/07/2012
« Reply #5 on: August 04, 2012, 02:26:58 PM »
Come on, you can do better than that. You have not even tried to check if your guess fits information given.
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #6 on: August 04, 2012, 02:46:42 PM »
Come on, you can do better than that. You have not even tried to check if your guess fits information given.

Negative. I actually tried and it didn't work  :'(. But I was not sure if I did it right as well. Now I know I did it right..so I won't post again until I have something that fits 100% with the info given- particularly the pressure, I can't think on something that fits with that.


Offline Borek

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Re: Problem of the week - 30/07/2012
« Reply #7 on: August 06, 2012, 02:23:25 PM »
Extended for another week.

You are right about HF reacting with SiO2. But HF is not the original gas.

What can you calculate using given information about weight?
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #8 on: August 06, 2012, 07:20:19 PM »
well - I can calculate the density of the gas

d=m/v= 10.2/1.22 = 8.36 g/L

Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #9 on: August 06, 2012, 09:09:03 PM »
okay- I finally found a gas that fits with:

-pressure change
-color

but still it doesn't fit with the density (I am comparing at room temp. and 1 atm pressure, from online tables)

if I find a gas that fits with the density---I'll post everything at once...

Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #10 on: August 06, 2012, 09:41:03 PM »

okay-I won't have time to continue with this problem for a few days, so I post what I tried if it helps to another person:


I tried with WF6, is a corrosive gas, density about 12 (that doesn't match) but it reacts with water to give:

WF6 + 3H20 = WO3 (yellow, dissolves in NaOH) + 6HF

Then HF produced reacts with SiO2:

4HF + SiO2 = SiF4 + 2H20

according to stoichiometry I got that, for each mol of WF6 that reacts, we get 3/2 moles SiF4 gas, that would give me the 50% increase in pressure.


 


Offline Borek

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Re: Problem of the week - 30/07/2012
« Reply #11 on: August 08, 2012, 02:15:58 PM »
density about 12 (that doesn't match)

Think it over.
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #12 on: August 08, 2012, 05:09:39 PM »
density about 12 (that doesn't match)

Think it over.

well since I saw the problem I wondered about at which pressure we were working, but I was afraid to ask since that could be a hint  ;D

I am thinking that if we are talking about a low pressure reaction vessel that can lower the gas density to 8.36 (not sure on this- I must say I don't remember much about under pressure vessels)
It's the only I can think of so far.

Offline Borek

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Re: Problem of the week - 30/07/2012
« Reply #13 on: August 13, 2012, 12:23:20 PM »
You can assume pressure around 1 atm.
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Offline Ann1234

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Re: Problem of the week - 30/07/2012
« Reply #14 on: August 13, 2012, 09:10:25 PM »
You can assume pressure around 1 atm.

ok so molar mass gas = 201.79 g/mol

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