# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on July 30, 2012, 01:01:00 PM

Title: Problem of the week - 30/07/2012
Post by: Borek 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.
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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???  ???
Title: Re: Problem of the week - 30/07/2012
Post by: Borek on August 01, 2012, 01:21:15 PM
Assume glass is mainly SiO2.

Nitrogen is not corrosive, it is almost completely inert.
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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)
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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)
Title: Re: Problem of the week - 30/07/2012
Post by: Borek 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.
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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.

Title: Re: Problem of the week - 30/07/2012
Post by: Borek 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?
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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...
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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.

Title: Re: Problem of the week - 30/07/2012
Post by: Borek on August 08, 2012, 02:15:58 PM
density about 12 (that doesn't match)

Think it over.
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 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.
Title: Re: Problem of the week - 30/07/2012
Post by: Borek on August 13, 2012, 12:23:20 PM
You can assume pressure around 1 atm.
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 on August 13, 2012, 09:10:25 PM
You can assume pressure around 1 atm.

ok so molar mass gas = 201.79 g/mol
Title: Re: Problem of the week - 30/07/2012
Post by: Borek on August 14, 2012, 04:17:14 AM
You can assume pressure around 1 atm.

ok so molar mass gas = 201.79 g/mol

No. But I am not able to reproduce the correct answer either  :'(, and as I calculated it in the Dalmatino bar using a napkin, I have no notes to consult.

Trick is, 10.2 g is not a weight of the added gas - it is the increase of the weight, so you should take buoyancy into account. As the volume is 1.22 L, and the air density is around 1.2 g/L, buoyancy is around 1.22*1.2=1.46 g - so the weight of the gas is around 10.2+1.4=11.6 g. That gives molar mass of around 230.

It should be 15.05-1.22*1.2=13.6 g, I am changing the wording of the original question. Sorry about that.
Title: Re: Problem of the week - 30/07/2012
Post by: Ann1234 on August 14, 2012, 12:31:26 PM
You can assume pressure around 1 atm.

ok so molar mass gas = 201.79 g/mol

No. But I am not able to reproduce the correct answer either  :'(, and as I calculated it in the Dalmatino bar using a napkin, I have no notes to consult.

Trick is, 10.2 g is not a weight of the added gas - it is the increase of the weight, so you should take buoyancy into account. As the volume is 1.22 L, and the air density is around 1.2 g/L, buoyancy is around 1.22*1.2=1.46 g - so the weight of the gas is around 10.2+1.4=11.6 g. That gives molar mass of around 230.

It should be 15.05-1.22*1.2=13.6 g, I am changing the wording of the original question. Sorry about that.

no worries  :)