1 when copper powder reacts with HCl acid at room temperature and pressure, the observations are as follows
Dark powder, changes to a pink powder. solution goes very pale blue. Explain this observation
2 0.004 moles of chromium reacts with 50cm3(an excess) of dilute hydrochloric acid at r.t.p. what is the charge on the chromium ion formed in the experiment? give reasons for your answer
3 Just for checking purposes, is chromium less reactive than aluminium?
that's all. thanks
Oh yea I just remembered another question!
goes something like
How does a soap manufacturer use the NaOH in making soap? name the process?
Originally posted by Audi:1 when copper powder reacts with HCl acid at room temperature and pressure, the observations are as follows
Dark powder, changes to a pink powder. solution goes very pale blue. Explain this observation
2 0.004 moles of chromium reacts with 50cm3(an excess) of dilute hydrochloric acid at r.t.p. what is the charge on the chromium ion formed in the experiment? give reasons for your answer
3 Just for checking purposes, is chromium less reactive than aluminium?
that's all. thanks
Cu powder black in colour should be CuO which reacts with acid to gives u Cu which is the pink powder which then reacts with HCl to give CuCl2 which is pale blue solution.
In a hurry now so no time to work out qn 2
Reactivity series check ur textbook, if its not inside then u should not need to know whether Cr is more or less reactive than Al unless they give u test results to deduce ur answers.
Solution : CuO2+ HCL ==> CuCL +H2O ( Copper chloride solution)
2. I remember that reaction forming Cr (III) ions, because that is more stable than the other chromium ions.
3. Aluminium is more reactive than Chromium.
Last one, NaOH is heated with fat to make soap. Called saponification.
Originally posted by dkcx:Cu powder black in colour should be CuO which reacts with acid to gives u Cu which is the pink powder which then reacts with HCl to give CuCl2 which is pale blue solution.
In a hurry now so no time to work out qn 2
Reactivity series check ur textbook, if its not inside then u should not need to know whether Cr is more or less reactive than Al unless they give u test results to deduce ur answers.
Hello! How does reacting CuO with HCl acid give you Copper? Shouldn't you get H2O and CuCl2 as stated by Bubblyl3?
Much thanks. Or is that a rapid process that occurs "momentarily"?
A company called CHLOROMAN makes chemical from rock salt (sodium chloride). NaCl arrives at the factory by pipeline as brine. CHLOROMAN sells the chlorine to a PVC manufaturer
If PVC manufacturer goes out of business, CHLOROMAN must find a new market for its chlorine AT ONCE. Why must CHLOROMAN do so? Suggest another buyer for which CHLOROMAN might sell its chlorine.
My answer isn't good but I'll tell you what I write so someone could improve on it... I wrote that chlorine, being a reactive halogen, will easily react after being put in place for a long time. So it has to find a new market. The buyer: Swimming pool? rofl.
Originally posted by Audi:
Hello! How does reacting CuO with HCl acid give you Copper? Shouldn't you get H2O and CuCl2 as stated by Bubblyl3?
Much thanks. Or is that a rapid process that occurs "momentarily"?
The question mentioned pink and the only thing pink is cupper metal. There could be a reduction of CuO to Cu as well as formation of the CuCl2 salt.
Originally posted by Audi:1 when copper powder reacts with HCl acid at room temperature and pressure, the observations are as follows
Dark powder, changes to a pink powder. solution goes very pale blue. Explain this observation
2 0.004 moles of chromium reacts with 50cm3 (an excess) of dilute hydrochloric acid at r.t.p. what is the charge on the chromium ion formed in the experiment? give reasons for your answer
(Not directed at anyone in particular) Don't automatically assume any question you read is valid, accurate, feasible, realistic or error-free.
"Copper powder" isn't "dark" or black, copper(II) oxide powder is. (FYI, copper(I) oxide is red).
The only scenario in which "Dark powder changes to a pink powder" may be observed, is when copper metal pieces (pink colour) are coated with a layer of copper(II) oxide (black colour); and when reacted with hydrochloric acid, the copper(II) oxide reacts to form soluble copper(II) chloride salt, which turns to solution pale blue, and reveals the (relatively unreactive with acids; depending on the specific acid and molarity used) pink copper metal underneath.
The chromium question is illogical, sloppy and may mislead/confuse students who read it. In similar questions in actual 'O' or 'A' level exams, the charge on the metal ion has to be calculated mathematically via redox stoichiometry and/or redox potentials. Giving the volume of 50cm3 is irrelevant or useless since such volume is stated to be in excess, and/or the molarity of the solution is not specified. The no. of moles of chromium is likewise irrelevant, since you're not given sufficient information to calculate the no. of electrons lost per mole of chromium. Basically all the numerical values given in the question are merely a useless bunch of red herrings, that has no bearing or relevance to the oxidation state of Cr in the ionic compound produced under reaction with acid.
Regarding the oxidation state of chromium when chromium metal reacts with hydrochloric acid, and additional information on chromium and its compounds in various oxidation states, visit :
http://en.wikipedia.org/wiki/Chromium
http://en.wikipedia.org/wiki/Chromium(II)_chloride
http://en.wikipedia.org/wiki/Chromium(III)_chloride
Originally posted by UltimaOnline:
(Not directed at anyone in particular) Don't automatically assume any question you read is valid, accurate, feasible, realistic or error-free.
"Copper powder" isn't "dark" or black, copper(II) oxide powder is. (FYI, copper(I) oxide is red).
The only scenario in which "Dark powder changes to a pink powder" may be observed, is when copper metal pieces (pink colour) are coated with a layer of copper(II) oxide (black colour); and when reacted with hydrochloric acid, the copper(II) oxide reacts to form soluble copper(II) chloride salt, which turns to solution pale blue, and reveals the (relatively unreactive with acids; depending on the specific acid and molarity used) pink copper metal underneath.
The chromium question is illogical, sloppy and may mislead/confuse students who read it. In similar questions in actual 'O' or 'A' level exams, the charge on the metal ion has to be calculated mathematically via redox stoichiometry and/or redox potentials. Giving the volume of 50cm3 is irrelevant or useless since such volume is stated to be in excess, and/or the molarity of the solution is not specified. The no. of moles of chromium is likewise irrelevant, since you're not given sufficient information to calculate the no. of electrons lost per mole of chromium. Basically all the numerical values given in the question are merely a useless bunch of red herrings, that has no bearing or relevance to the oxidation state of Cr in the ionic compound produced under reaction with acid.
Regarding the oxidation state of chromium when chromium metal reacts with hydrochloric acid, and additional information on chromium and its compounds in various oxidation states, visit :
http://en.wikipedia.org/wiki/Chromium
http://en.wikipedia.org/wiki/Chromium(II)_chloride
I think i remember doing 1 question in uni in which either Cu2O or CuO was reacted with H2SO4 i think and the resultant give both a Cu salt and Cu metal
Originally posted by dkcx:I think i remember doing 1 question in uni in which either Cu2O or CuO was reacted with H2SO4 i think and the resultant give both a Cu salt and Cu metal
H+ and SO4 2- are both oxidizing in nature, not reducing. Can you try to recall what you did, and write the half-equations and balanced redox equation to show how Cu metal is obtained?
A possibility (assuming copper metal is indeed produced), would be that as the oxide anion is protonated by the acid, this 'frees up' the copper cation to be reduced by the water solvent. This scenario would be more likely for Cu+ ion than the Cu2+ ion, as the fomer has a larger reduction potential than the latter.
Can you locate any reliable references to support such ideas of copper metal produced when a copper oxide is reacted with acid?
Originally posted by UltimaOnline:
H+ and SO4 2- are both oxidizing in nature, not reducing. Can you try to recall what you did, and write the half-equations and balanced redox equation to show how Cu metal is obtained?
A possibility (assuming copper metal is indeed produced), would be that as the oxide anion is protonated by the acid, this 'frees up' the copper cation to be reduced by the water solvent. This scenario would be more likely for Cu+ ion than the Cu2+ ion, as the fomer has a larger reduction potential than the latter.
Can you locate any reliable references to support such ideas of copper metal produced when a copper oxide is reacted with acid?
Here is the question. Don't have the answers or should i say i'm abit lazy to search through all my yr 1 notes to look for this tutorial answer. Maybe you can try work it out but our prof said it was possible for Cu metal to be produced when we questioned it and we don't really argue back about such stuff.
(3) Solutions of CuSO4 and 2 equiv NaOH react to give a light blue precipitate Cu(OH)2. 0.499 g of Cu(OH)2 is heated at 1000 oC to give a red solid A that weighs 0.316 g. This red solid A is an oxide of cupper. When A is dissolved in H2SO4 solution, another red solid in the elemental form that weighs 0.141 g is obtained.
i) Find out the (empirical) formula of A. Show your result based on calculations.
ii) White all the chemical equations involved.
Originally posted by dkcx:Here is the question. Don't have the answers or should i say i'm abit lazy to search through all my yr 1 notes to look for this tutorial answer. Maybe you can try work it out but our prof said it was possible for Cu metal to be produced when we questioned it and we don't really argue back about such stuff.
(3) Solutions of CuSO4 and 2 equiv NaOH react to give a light blue precipitate Cu(OH)2. 0.499 g of Cu(OH)2 is heated at 1000 oC to give a red solid A that weighs 0.316 g. This red solid A is an oxide of cupper. When A is dissolved in H2SO4 solution, another red solid in the elemental form that weighs 0.141 g is obtained.
i) Find out the (empirical) formula of A. Show your result based on calculations.
ii) White all the chemical equations involved.
Indeed as I predicted based on reduction potentials, it is Cu+ rather than Cu2+ that is reduced. This is less likely to occur with a Cu2+ salt (consider their different reduction potentials). By Le Chatelier's principle, the greater molarity of water as a solvent (as I've already earlier mentioned, the reducing agent is unlikely to be the protons or the sulfate(VI) ions from the sulfuric(VI) acid, which are both oxidizing in nature) enables the reduction of Cu+ ions to elemental Cu to be feasible. The trick here, is that the opportunity for such to occur, is presented by the protonation of oxide anion.
Regardless, most things are possible, but under specific circumstances. Just because a question implied a theoretical possibility, does not necessarily mean it's practically feasible under normal circumstances.
Good discussion.
very chim...
Originally posted by UltimaOnline:
Indeed as I predicted based on reduction potentials, it is Cu+ rather than Cu2+ that is reduced. This is less likely to occur with a Cu2+ salt (consider their different reduction potentials). By Le Chatelier's principle, the greater molarity of water as a solvent (as I've already earlier mentioned, the reducing agent is unlikely to be the protons or the sulfate(VI) ions from the sulfuric(VI) acid, which are both oxidizing in nature) enables the reduction of Cu+ ions to elemental Cu to be feasible. The trick here, is that the opportunity for such to occur, is presented by the protonation of oxide anion.
Regardless, most things are possible, but under specific circumstances. Just because a question implied a theoretical possibility, does not necessarily mean it's practically feasible under normal circumstances.
Good discussion.
I would say at O's or A's level, most students won't be questioning the conditions of reactions nor understand the conditions required but will just plainly accept the question as its given. There is no time to argue with the teacher in the exam that the question is not practical and has flaws which makes the answer impractical etc