if Mg3N2 is not completely decomposed, will experimental mole ratio of Mg to O be too low or too high?
823593
When magnesium reacts with oxygen gas: \n 2 Mg  +  O2   ---> 2 (MgO)  the ratio of Mg to O is 1:1\n(1 mole of Mg reacted produces 40.3 g of MgO)\n\nWhen it reacts with nitrogen gas\n 3 Mg +   N2   --->  Mg3N2  the ratio of  Mg to N is 1: 0.67\n(1 mole of Mg reacted gives 33.6 g of Mg3N2)\n\nSo if some Mg3N2 gets left behind in what you think is pure MgO, then the mass will be lower than it should be. Your calculated mass of oxygen reacted will be lower than it should be, too.\n\nWhen you convert to moles, you will have fewer moles of oxygen \nthan you should have,  i.e., the ratio of Mg to O will be too high.
728971
(The) second experiment was to observe the reaction when a magnesium ribbon was ignited. The ribbon was put to the flame of a burner (delete ,) and was ignited. The reaction was very (intensive) (change to intense), and (burned with a) (delete was with) very bright white light. Usually when a substance is ignited in air, it reacts with H2 or O. In this case, because the Mg was the positive ion, it must have reacted with oxygen, (which is a ?)negative ion. In this experiment, because Mg was joined to O2 to make 2MgO of different composition, this was a combination reaction.\nThe third experiment 0.1 M CuSO4 and 6 M NH3 were used. CuSO3, a Sky blue liquid,  was place in a test tube, and then 1 drop of 6 M NH3 was added. Right after NH3 was added, the top of (the) liquid changed to navy color.  After the liquid was stirred the color of the liquid changed to a blue that was darker than the color of CuSO3. Ten more drops (of CuSO4?) were added to the contents, and the color of (the) contents changed to a violet color.
940335
In your crucible, you have a mixture of mostly MgO and some Mg3N2.  That combined mass is 0.470 g.\n\nWhen you add water, all you are really adding is oxygen, since the hydrogen leaves as part of the ammonia gas.\n\nBasically, you are adding oxygen (atomic weight 16) and removing nitrogen (atomic weight 14).\n\nBecause of the formula Mg3N2(s) + H2O(l) -> MgO(s) + NH3(g),\nyou are 'taking out' 2 nitrogen atoms for every 3 oxygens you add (because there are 2 nitrogens to every 3 magnesium atoms, and each magnesium atom uses 1 oxygen to make MgO)\n\nSo you are gaining (0.486 - 0.470) = 0.016 g from the reaction of Mg3N2 to MgO\n\nThe mass from adding the oxygen is 16 g/mol * (3) * X\nThe mass from taking out the nitrogen is -14 g/mol * (2) * X\n\nWhere X is the number of moles of Mg3N2\n\nSo, 16 g * 3 * X - 14 g * 2 * X = 0.016 g\n\nSolving for X, X = 0.016 / (48 - 28) = 0.0008 mol\n\nThe molecular weight of the Mg3N2 is 100.9 g/mol, so the mass of Mg3N2 is 100.9 g/mol * 0.0008 mol = 0.0807 g\n\nSo, the mass percent is .0807 / 0.470 = 17.2 % Mg3N2\n\n.............\n\nAn another note, one reason you might be confused is your second reaction:\n\nThere are actually 2 reactions going on:\nMg + 1/2 O2 -> MgO\n3Mg + N2 ->Mg3N2\n\nThe first reaction reacts much faster than the second, meaning that more MgO will be produced than Mg3N2.\n\nHope this helps.
819664
The lone pair of electrons on the Nitrogen atom of Ammonia, is available for donation. That is why ammonia is considered as a base. \nBut it not as readily available as it is From OH- radical or other basic radicals. So it is much weaker than them.\nIn aq. soln, NH3 combines with water to form a very weak electrolyte NH4OH, which gets dissociated to a very small extent furnishing only a few OH- ions. So the aq. soln. of NH3 is also mild basic.\n\n Ammonia acts both as a weak base in the Brernsted-Lowry sense\nNH3 + H2O = NH4(+) + OH(-)and\nas a complexing agent (Lewis acid) with some metal ions\nZn(2+)+ 4NH 3 + Zn(NH3)4(2+)\nBoth weak and strong bases produce hydroxide ion in solution which precipitates some metal ions, e.g.\nA13(+) + 30H- + Al(OH)3(s)\nand Zn92+) + 20H- + Zn(OH)2(s).\nThese precipitates appear as flocculent white solids which settle slowly from solution. In higher\nconcentrations hydroxide ion acts as a complexing agent for amphoteric cations, e.g.\nZn(OH)2(s) + 2OH- -+ Zn(OH),2-\nAl(OH)3(s) + OH- -+ A1(OH)4-\nThis experiment demonstrates the action of OH- and NH3 on Al”(aq) and Zn*‘(aq) ions.\n\nAlthough ammonia is well-known as a base, it can also act as an extremely weak acid. It is a protic substance, and is capable of dissociation into the amide (NH2−) ion, for example when solid lithium nitride is added to liquid ammonia, forming a lithium amide solution:\n\n        Li3N(s)+ 2 NH3 (l) → 3 Li+(am) + 3 NH2−(am)\n\nThis is a Brønsted-Lowry acid-base reaction in which ammonia is acting as an acid.\n\nI think you are interested about dry ammonia.
1141237
depends if it is 25.9% nitrogen by molar weight or total # of atoms
659725
Magnesium will react with any gas except noble gases when heated, because of its low electronegativity. In vacuum or noble gas atmosphere it can be heated up to melting point of 625°C, just at or below red glow.\n\nWhile red or white hot iron will oxidize on the surface, the heat released by that process is not sufficient to start a chain reaction like in burning magnesium. The higher melting point of iron depends mostly on the crystal lattice of iron.
708251
You should expect a molecular compound to form between N and O (as opposed to an ionic compound).  This is because N and O are both nonmetals.  An ionic compound will form when a metal and a nonmetal combine.  I believe that this is what your question is getting at.
194284
To solve this problem we need to think about limiting reactants.  This one is fairly simple because I already have moles and the ratio is 1:1:1.  If I only have 3 kmols of C, that is the most I can make, because going beyond that, I would be out of C.  Thus, I can only make 3kmols of CO2 with 2 kmols of O2 left unreacted.  These amounts are theoretically find for a combustion reaction.
249344
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.\n\nWe solve this problem by first writing the equation\nN2 + 3H2 = 2NH3\nN2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g\nso 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent.
644051
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
951269
For all practical purposes, if a solution is saturated (can not dissolve any more) it is full and all others will fall to the bottom, undissolved. \n\nRealistically a small (very small) amount of the other powder could dissolve if it had different enough properties.
794643
1. highter since given weight of magnesium looks to be a larger percent of finished product.\n2.It would depend on molecular weight of other reactants.  If they were heavier than Oxygen the Mg would seem to be a lower percentage.  If only other possibility for Rx was with nitrogen and ratio of recatants was the same as with Oxygen(unlikely) then the resulting compound would be lighter than MgO making the Mg percent higher.\n3.  Mg oxydizes in air at room temp.  This white film was MgO already oxydized which would interfere with your results, raising the calculated percentage of Mg.
480464
YOU WOULD HAVE TO USE THE BOND ENTHALPIES TO WORK OUT WHICH ONE. SEEING THOUGH N2 MAKES UP 78%OF THE EARTHS ATMOSPHERE ITS SEEMS VERY INERT SO MOST PEOPLE WOULD ASSUME THAT IS THE ONE
798796
Not impossible but Hco3- is considered a very weak acid with resilience towards losing another Hydrogen Ion that's a good explanation for Hydrogen Carbonate formation instead of Carbonate
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
878297
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?
663581
NH4NO2 --------------> N2 + 2H20 \n\nYOU NEED TO HEAT THE SOLUTION IN ORDER TO DECOMPOSE THE COMPOUND \n \nHOPE THIS HELPS