Small quantities of oxygen gas are sometimes generated by heating   KClO3   in the presence of   MnO2 HELP?
960502
Since the oxygen is collected over water, you must subtract out the water vapor pressure to determine the pressure due to just the oxygen.\n\n722.2 mmHg - 21.1 mmHg = 701.1 mmHg = 0.922 atm due to O2\n\n0.276g KClO3 = 0.00175 moles KClO3 (molar mass = 158 g/mol)\nSince KClO3 decomposes to form O2 in a 2:3 ratio, the number of moles of O2 produced = 0.00262 moles O2\n\nPV=nRT\n(0.922 atm) V  = 0.00262 mol (0.08206)(296.2K)\n\nSolving for the volume due to the O2,  V = 691 mL
673183
MnO4- + 5e- + 8H+ -> Mn2+ + 4H2O\n\nVO+2  + H2O ->  VO2+ + e- + 2H+\n\nMultiplying eqn. below by 5\n\nMnO4- + 5VO+2 + H2O -> Mn2+ + 5VO2+ + 2H+
224650
Atoms favor complete valence shell fulfillment and charge neutrality above all.  Since potassium has a positive center that requires 1 electron above the more stable electron orbital fulfillment, it does not require much energy to release the electron.  Since our environment consists of O2, it will react to it because a more stable configuration than the O=O configuration is the K2O.  Why?  Oxygen's center does not have enough electrons to complete the p orbital shell.  Therefore, it is not satisfied.  Thus, oxygen will share 2 electrons with another oxygen to form O2 until it has an opportunity to take 2 K electrons that potassium does not want.  Therefore, Oxygen is a lot happier than it was with another Oxygen atom who wanted the 2 electrons.  Potassium is a lot happier getting rid of the electron that it did not want.  But wait?  Why are they stuck together?  Because despited wanted to shape their electron clouds at the most ideals shapes possible, the nucleus still requires an equal amount of electrons closeby to be happy.  Thus, the only way to achieve this is to stick around Oxygen who's slightly negative.\n\nThe same could be said between Potassium and any halide, but not Potassium and any of the noble gases.  Noble gases are really happy to be complete because their electron configurations are at an ideal state - completely fulfilled.  To strip an electron would require a really high amount of energy.\n\nThus, the reasons why K likes most stuff that aren't satisfied, but not all.
158393
You could use hot KMn04, but what about using a chromium oxidizing agent, such as Cr03 or NaCr04 ... you might get a cleaner oxidation.
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
363702
changes oxygen into carbon dioxide, the carbon come from the burning fuel
446321
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
336167
oxygen and carbon dioxide
74307
The permanganate is reduced to Mn2+ and the oxalate is oxidized to CO2.\n\n2 MnO4- +  5 C2O42-  +  16 H+ =  2 Mn2+  +  10 CO2  +  8 H2O\n\nYou obviously know enough to ask the question so I'm assuming you can maneuver the numeric placements appropriately.  This forum does not allow me to sub or super set characters.
561048
Some oxide of chlorine. No idea about which one would be stable at those conditions, though.
982213
PV = nRT\n\nP = 728-21.1 / 760 = .93 atmospheres\nV = unknown\nn= .281 g KClO3 x mol KClO3 / 122.5 g x 3 mol O2 / 2 mol KClO3 = .003441 moles O2\nR= a constant = .08206\nT= 23 deg celsius + 273 = 296 Kelvin\n\nV = (.003441 x .08206 x 296) / .93 = .0899 L = 89.9 mL
12446
Both group 1 and 2 nitrates give oxides on heating.\ngroup 1 carbonates are thermally very stable but group 2 carbonates are thermally unstable, the unstability increasing with cationic size.
929210
You are exactly right. The energy given to the system when you hit them together was enough energy to break the bonds of the oxide layers on top. The thermal energy of the reaction elevated the electron energy states for a small amount of time, causing them to fall back down to their ground state and giving off some photons...hence the spark.
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
175333
I'm sure the KCl02 --> to KCl and CO2.\nThe clorate likes to give up the oxygen so look up the bluestone and see if it will react with O.\nThe first is a chemical reaction.\nmaybe the bluestone also.
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
867182
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O