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Making H2SO4 at home is extremely difficult, if not impossible. Making it from HCl is not possible, because on heating, the HCl always is driven away. HCl is a gas and easily escapes the mix. The only possible way would be to take HCl and a soluble metal sulfate, which gives an insoluble metal chloride. There are only few metals, capable of this. The sulfate needs to be soluble, while the chloride is not. Ag(+) and Tl(+) have this property, but none of these you'll probably use for making H2SO4. Otherwise, forget about making the H2SO4. You could have some luck at automobile stores and garages, where they sell battery acid. That is somewhere between 30% and 40% H2SO4, which is quite pure. You could concentrate that to 90+ %, but that is a very dangerous operation, which only should be carried out with the proper equipment (glassware) and fume hood (or outside).

Dissolve the magnet in concentrated HNO3 or in aqua regia. The resulting solution certainly is not magnetic anymore .

You should not look at the temperature drop after a certain time, but at the temperature drop, after a certain amount has evaporated. So, if you wait, until e.g. 5 grams are evaporated, then I expect that with heptane the temperature drop is larger than with hexane. What you did, is measure temperatures every few seconds (minutes?) and compare these. That is wrong. Less heptane evaporates per unit of time than hexane. Apparently the effect of slower evaporation outbalances (and in fact overrides) the effect of more energy consumption per gram of compound. Repeat the experiment with two balances (accurate) and start with e.g. 10 gram of each liquid. Each time, when e.g. 0.1 gram is evaporated, measure the temperature. Now you'll have graphs of evaporated mass vs. temperature. If you make graphs for both compounds, then I expect that the heptane case shows lower temperatures. When performing this experiment, however, you have to take into account that the liquids, cooling down due to evaporation, also receive heat from the surrounding air. That may spoil the results somewhat, this effect will be stronger in the heptane case.

Such a liquid does not exist. You can, however, prepare a mix of CCl4 and CS2. That mix burns at fairly low temperature, low enough to pour some of this on your hand and then ignite it. You can stand its temperature for a while, although not indefinitely. A serious drawback of this combo is its extreme toxicity. CS2 is a strong nerve toxin, and CCl4 is highly carcinogenic and it kills your kidneys (and probably you as well). In the past this was used in some magicians shows, but that practice has been abondoned, because of the toxicity of this mix.

HCl (or any other other acid) only reacts with oxides at an appreciable rate if they are not calcined or crystalline. HCl indeed reacts (slowly) with fresh rust (hydrous Fe2O3) or freshly precipitated Fe(OH)3. The solutions formed this way are not green, but bright yellow, due to formation of the complex ion FeCl4(-). Reaction with Fe3O4 probably is nill, because that also is a very compact, waterfree oxide. This effect is quite common. Many freshly precipitated metal hydroxides or oxides easily dissolve in HCl or dilute H2SO4, but aged and dried oxides, or calcined oxides, usually do not dissolve anymore and are remarkably inert. Examples of these are Cr2O3, Al2O3, TiO2, Fe2O3, Nb2O5, Co3O4.

@budullewraagh: If I look at that bottle, I think you have a solution of I3(-) ions. Such a solution gives off iodine vapor very slowly and this vapor probably escapes through the cap slowly, causing stains on all kinds of things nearby. I myself also have a bottle of a solution of KI3, made by dissolving I2 in concentrated solution of KI. This bottle looks very similar. The liquid has a strong smell of iodine. @Jdurg: I would write [i.I2](-) for the triiodide ion and [Cl.ICl3](-) for the ICl4(-) ion, and [Cl.ICl](-) for the ICl2(-) ion. These formulas best represent the structure of these ions.

Yes, but it is not the best way to make HNO3. Don't you think it is a pity to use your copper nitrate for that? How much HNO3 and at what concentration do you want to make? Making HNO3 of a concentration above let's say 40% will be hard. At higher concentration, the NO2 does not react any more with the water and NO and NO2 simply dissolve. The liquid will turn dirty green, due to a mix of red/brown dissolved NO2 and blue N2O3. So, you'll end up with a liquid containing water, HNO3, NO2 and N2O3 (in equilibrium with NO+NO2). If you want to make HNO3 at a larger scale, then I would go for distillation, simply at air pressure of a mix of KNO3 (or NaNO3) and conc. H2SO4, with a little water added. For details, YT can give better advice. For me, making HNO3 is not useful at all, I can simply buy 52% HNO3 for just a few bucks per liter, but I know it is different in many other countries.

In the I3(-) ion, the three iodine atoms are not all equivalent, in fact, one of them can be regarded as having oxidation state -1, while the others have oxidation state 0. This is quite different from azide, N3(-), where indeed all N-atoms are similar and each N atom truly has oxidation state -⅓. If you have a solution of KI and Cl2 is bubbled through, then you first get I3(-) ions. Continued bubbling yields ICl2(-) ions (orange) and even further bubbling yields yellow ICl4(-) ions. The latter ions are much more stable and it is very difficult to isolate ICl2(-) ions, e.g. as KICl2. The precise compound formed also depends on concentration and pH. At high pH, these polyhalide ions are unstable and instead, iodate ion is formed, IO3(-). At intermediate or low pH, at low concentration, I2 and Cl(-) are formed. When more chlorine is bubbled in, IO3(-), H(+) and Cl(-) are formed. Only at very high concentration, the polyhalide ions ICl2(-) and ICl4(-) are formed. In fact, the chemistry of these polyhalide ions is very subtle and sensitive to the reaction conditions.

Cu(NO3)2.3H2O decomposes at a fairly low temperature. It gives NO2, O2, H2O and probably also some NO on decomposition. What remains behind is CuO. The wikipedia page is correct. First, KNO3 decomposes at around 400 C to give KNO2 and O2. This reaction is very slow. Getting NOx from KNO3/KNO2, however, requires much higher temperatures, somewhere around 1100 C.

It is hard to say whether an explosion will occur in that case. I don't expect so, but there are a few factors, which could cause "explosion-like" reaction. 1) Temperature rise, due to exothermic reaction 2) Liquids becoming a gas again due to heating up 3) Formation of additional water, causing even faster reaction. In practice I think that you will see a violent, but not explosive reaction, unless you confine the reaction mixture to a small space. Mixing liquid SO2 and H2S seems a dangerous experiment to me anyway. Even traces of water can initiate a reaction. You must be absolutely sure that the material is REALLY dry.

You need to know the lattice energy of Na2O, that is the energy, released when two moles of Na(+) ions and one mole of O(2-) ions (theoretically) are combined to the solid Na2O. This energy is given in kJ/mol. I don'h have this info at hand, but there are tables for lattice energies of many ionic compounds.

Ionization takes a lot of energy, so that would indeed mean that the reaction is very strongly endothermic. But... you forget about the crystal lattice formation. The combining of O(2-) ions and Na(+) ions gives a lot of energy, so much, that the net reaction is strongly exothermic and that is exactly what we observe. Na metal burns very well (too well, actually).

But keep in mind, that at high pH a totally different reaction occurs. One of the products is CHI3, the other is acetate ion. Look at "haloform reaction" in google. This will certain help you.

If you can get NO2, then you can make HNO3. But heating NaNO3 or KNO3 in order to make NO2 is VERY hard. You cannot accomplish that at home, the tempetatures needed for that are really high (IIRC over 1100 C). First, you get loss of O2, forming KNO2 or NaNO2, but this process also requires very high temperatures and cannot be done at home. What is feasible, however, is to make NaNO2 or KNO2 from molten KNO3 and adding lead metal. KNO3 + Pb --> KNO2 + Pb Use excess KNO3 and add the metal to the molten KNO3. The reaction is smooth, not violent. Let the melt solidify. No need to purify it. Add dilute HCl to the solid PbO/KNO3/KNO2 mix. You'll get loads of brown fumes (NO + NO2). These should be bubbled through water. You also need to assure that fresh air can enter the liquid, otherwise you loose a large part of your oxides (NO does not dissolve in water, but reacts with O2 from air to form NO2, which dissolves in water). I don't think this is a really efficient route to HNO3, but it can be used to make crude dilute HNO3. If you perform this synth, be very careful with the molten KNO3-mix and also be very careful with the insidiously toxic nitrogen oxides. These can kill you. Keep in mind, that NO2 has delayed effect. You can feel quite well immediately after breathing a lot of NO2, while 24 to 48 hours later you suffocate, due to buildup of water in your lungs (IIRC this is called lung oedemia (sp?)). Be warned, this is horrible and a friend of mine has experienced this to quite some extent after inhaling too much NO2! He felt really bad for more than one week, but fortunately he had no irreversible damage!!

Yes, that is correct. In this situation, the reactants match precisely, so nothing remains unreacted. You will get a limiting reagent if you change the ratio, e.g. 0.003 l of Pb(NO3)2 and 0.005 l of Na2CO3. Just do the math and determine, which is the limiting reagent, how many moles of the reaction product you get and how many moles of the other reagent remains. As another exercise for you to see if you really understand. Replace the solution of 0.5M Na2CO3 by a solution of 0.5M NaI. With NaI you also get a similar reaction: Pb(NO3)2 + 2NaI --> PbI2 + 2NaNO3 Here PbI2 precipitates from solution (a yellow precipitate). Again, assume total volume is 0.008 l. Determine how much solution of Pb(NO3)2 and how much solution of NaI is needed to have precisely no limiting reagent.

I don't know of any fast reaction between iodine and propanone at low pH. Maybe that boiling gives some reaction, but still I don't expect any. Increase the pH to 13 or 14 and then you'll see a fast reaction, yellow iodoform, CHI3, is formed in that reaction, which quickly settles at the bottom and can easily be isolated.

What reaction do you expect between the iodine and the propanone? I think you need a LOT of patience for this reaction . Propanone (=acetone) and iodine do react, but only at high pH. Then you get the so-called haloform reaction, yielding iodide and iodoform, CHI3. The haloform reaction is quite fast.

Indeed, you have to do what your friend suggested. The TOTAL volume is 0.008 liter, but the volume of Pb(NO3)2 solution and the volume of Na2CO3 solution is lower (e.g. 2 ml and 6 ml). So, you compute how many moles of Pb(NO3)2 you have from the original volume of that solution and you compute how many moles of Na2CO3 you have from the original volume of solution of Na2CO3. Now, when you have the number of moles for both compounds, you can determine, which is the limiting reagent and then you can compute the amount of PbCO3 formed in the reaction. The second question is not clear to me. What do you mean with "precipitate height"? Is that the total height of the liquid? The only way to measure the precipitate correctly is to filter it out of the solution, rinse it carefully and dry it carefully, without loosing any precipitate and weighing the amount of precipitate. If with measuring the precipitate height you mean measuring the height of the spongy or jelly like structure in the liquid, then I think that you are doing a useless thing. The density of the precipitate may vary wildly.

If what you are saying is the case, then I would strongly complain at the school director or whatever appropriate place. Without basic understanding of the concept of stoichiometry this kind of exercises cannot be solved. The idea is that at both sides of the arrow (see post of Tom Mattson), the number of atoms for each involved element must be the same. For carbon, this means that x = b For oxygen this means that z + 2a = 2b + c For hydrogen this means that y = 2c Another constraint is that the total weight at both sides of the arrow must be the same. Using atomic masses for the elements C,H,O, this gives another set of equations in x,y,z and a,b,c. Solving this set of equations will give you the answer. EDIT: @Tom Mattson: While I was typing, you just added your reply. But, indeed, I totally agree with you .

That might quite well be possible. I noticed a sweet smell from the brown liquid. I concluded that this is some chlorated hydrocarbon (the ones I have, have a sweet smell to me, CCl4, CHCl3, CH2Cl2, CHCl=CCl2), but I did not immediately conclude that it is CHCl3.

Indeed, as Jdurg states, gases, once mixed, will not separate anymore. It is good that they do not, otherwise we would have a bad time here on earth with the mix of gases in the atmosphere . If you perform electrolysis, then assure that the H2 and Cl2 are collected in different bottles. Never let them bubble into the same container. If you start correctly, then you'll not have the problem of separation at all.

This is a nice experiment, which shows formation of the special salt KICl4, which is a member of a whole class of very reactive and unstable salts (e.g. KI3, KIBr2, CsICl2, KICl4). http://woelen.scheikunde.net/science/chem/exps/KIO3+HCl/index.html You only need fairly common reagents, which are not that hard to obtain.

Glucose: CH2(OH)CH(OH)...CH(OH)C(O)H Assume oxidation state of H equals +1 (not -1, because H is a stronger reductor than C), oxygen equals -2. The last C is in a more oxidized state (it has an aldehyde group on it), than the other 5 C's. The first C has two H's and one -OH and hence is in a more reduced state than the other C's. So, oxidation state of first C equals -1, oxidation state of middle four C's equals 0, and the oxidation state of the last C equals +1. Glucose also exists in a ring-form (in reality there is an equilibrium between these two forms). I do not precisely recall the ring form, but most likely in that case, the oxidation state of the two end-C's will be different, but in order to determine, one has to checkup the ring structure.