[valyr]

I have an inorganic question but it seems more like high school chemistry. I'm trying to identify the substances used in the following scheme of reactions designed to figure out the reactivities of the group 16 elements.

a.) A metal (A) reacts with water to give a colorless solution of compound (B) and a colorless gas (C). A common dilute diprotic acid (D) is added to (B), forming a dense white precipitate (E).

b) A solution of (F) slowly decomposes to give a liquid (G) and a colorless gas (H). Gas (H) reacts with colorless gas (C) to give liquid (G).

c) Under certain conditions, colorless acidic gas (I), will react with gas (H) to give a white solid (J). Addition of (G) to (J) gives a solution of acid (D).

d.) Metal (A) burns in excess gas (H) to give a compound (k). Compound (K) dissolves in water to produce a solution of (B) and (F).

My first guesses for (A) were Ca, Ba, and Sr. Of the 3, Ba made the most sense since Ba(OH)₂ is colorless and Ba is fairly reactive enough to evolve hydrogen as from water. Ca(OH)₂ isn't colorless at all (lime water) and Sr(OH)₂ is already colored at that point. I figured (C) would be just hydrogen gas. (D) seemed to be sulfuric acid, common and diprotic. upon addition to barium hydroxide, barium sulfate is formed, which is a dense white precipitate (E).

The problem lies from compounds F to K. I figured (F) would be carbonic acid, but that decomposes far too fast.

Compounds I and H together yield J. But G and J yield D, which is sulfuric acid. So either G or J must contain some sort of sulfate. I assume it can't be G, because I'm not aware of any slowly decomposing solutions containing sulfur. It could be J, since I is a colorless acidic gas, so I'm thinking (I) could be Hydrogen sulfide. But then again, I'm unaware of any sort of reaction having sulfur yield a white solid.

I have a feeling compound K is plain ol' Barium Oxide, since Barium does burn in Oxygen to give BaO. and BaO dissolves in water to give Barium Hydroxide. Unfortunately, it seems to be wrong as two compounds are formed, B and F. And F, I am unsure of.

It is also possible, A to E is wrong.

So uh, any help/advice/hints?

[Astrokel]

Hey,

my thought to this question is to start off with the assumption of D as H2SO4 (as you suggested) and colourless acidic gas I as SO2 since this question is on group 16 elements as well as sulphur compound being involved.

Under certain conditions as you have said, I, SO2 will react with H to form J.

SO2 + 1/2O2 ---> SO3 (well known contact process, happened under certain conditions)

Therefore H is O2 and J is SO3

Im pretty sure A is barium and B is barium hydroxide and thus C is hydrogen gas.

Therefore H + C ---> G

G is H2O since H is O2 and C is H2

D + B  ---> E

E has a high chance to be BaSO4

G + J ---> D also makes sense now at this point,

G is water reacts with SO3 to form H2SO4, contact process!

A + H ---> K

Ba + O2 ---> BaO2, barium peroxide

F ----> G(H2O) and H (O2)

Therefore the only possibility is F being hydrogen peroxide, H2O2

and lastly,

K dissolves in water ---> B + F

BaO2 + 2H2O  ---> Ba(OH)2 + H2O2


Hence,

A - Ba
B - Ba(OH)2
C - H2
D - H2SO4
E - BaSO4
F - H2O2
G - H2O
H - O2
I - SO2
J - SO3
K - BaO2


My only worry is SO3 being a white solid at certain condition?

 

[Borek]

My only worry is SO3 being a white solid at certain condition?

http://en.wikipedia.org/wiki/Sulfur_trioxide

While I understand your joy after you have solved the riddle, giving answer out immediately was not the best idea. Please remember about forum rules in such situations.

[Astrokel]

My only worry is SO3 being a white solid at certain condition?

http://en.wikipedia.org/wiki/Sulfur_trioxide

While I understand your joy after you have solved the riddle, giving answer out immediately was not the best idea. Please remember about forum rules in such situations.

Very sorry about it  Or it is possible to edit the post for me, so that it wouldn't be so direct, sorry for any incovenience caused.

http://web1.caryacademy.org/chemistry/rushin/StudentProjects/CompoundWebSites/2000/HydrogenPeroxide/history.htm

and SO3 is a gas in the contact process, how can it be a white solid 

[valyr]

I must say thank you very much, although I did not expect the entire answer outright.

the material safety data sheet shows sulfur trioxide as a solid (bp at 44.7 C), so i'll take it's word for it.

i would not have suspected hydrogen peroxide to be the mystery decomposing liquid, since i expected it to decompose as fast as, well, carbonic acid. the contact process is also referenced in my textbook, so that seems to check as well.

So I take it plain BaO does not make H₂O₂ in solution? That extra oxygen in BaO₂ makes all the difference? It seems like I'm much more likely to get BaO from burning Ba in O₂ than BaO₂

[Astrokel]

well my bad,

both strontium and barium form peroxides upon heating with oxygen, the difference is strontium only forms peroxide at high pressure, whereas barium forms a mixture of oxide and peroxide upon heating with oxgen.