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1. An unknown oxide of mercury decomposes when heated to form mercury metal and oxygen gas (O 2). When a 1.048 g sample of this unknown is heated, 0.971 g of mercury remains. a. Calculate the moles of mercury and moles of oxygen in the compound. b. What is the empirical formula of this oxide? 2. a) A side reaction in today’s experiment occurs between some of the magnesium and the nitrogen gas, N 2. Write a balanced chemical equation for this reaction. b) The magnesium nitride can be converted to magnesium oxide by the addition of water. Write a balanced chemical equation for this reaction. 3. Suppose 1.087g of magnesium is heated in air. What is the theoretical amount of magnesium oxide that should be produced? 4. Calculate the percent by mass of molybdenum and sulfur in Mo2S5. Hi, Am I on the right track with this first part? Thanks |
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| | From:  ·Steve· | Sent: 1/18/2006 7:15 AM |
Right method, except the original compound is an oxide of mercury, not magnesium, HgxOy. That's why you got such a large ratio of "Mg" to O, because the atomic weight of magnesium is so much less than that of mercury. X and Y are the simplest whole numbers of moles of mercury atoms relative to oxygen atoms, so if you calculate the moles of O2 molecules that are produced (as you did, dividing the mass of oxygen by 32.00 g/mol) don't forget to multiply this result by 2 mol O / mol O2. Or, you can just divide the mass of oxygen formed by the atomic weight of oxygen, 16.00 g/mol, to get the moles of oxygen atoms. It is true that O2 molecules are actually formed, but it is the moles of O atoms that we need for the purpose of determining the empirical formula.
Steve
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|  0 recommendations | Message 3 of 11 in Discussion |
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How does this look for the first part? |
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Then for these, how does it look (pic)? Thanks, Steve!! 2. a) A side reaction in today’s experiment occurs between some of the magnesium and the nitrogen gas, N2. Write a balanced chemical equation for this reaction. b) The magnesium nitride can be converted to magnesium oxide by the addition of water. Write a balanced chemical equation for this reaction. 3. Suppose 1.087g of magnesium is heated in air. What is the theoretical amount of magnesium oxide that should be produced? 4. Calculate the percent by mass of molybdenum and sulfur in Mo2S5. |
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Thanks again Steve - you have a sharp eye! I am going to ask about: 2b. The products are Mg(OH)2 (s) and NH3 (g). You can smell the ammonia if put your nose close to the crucible and sniff cautiously immediately after you add the water. The Mg(OH)2 is converted to MgO (s) + H2O (g) by reheating. Looks like this question was misworded, as MgO is not the immediate product when the water is added. I think they need to attend your class.  |
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| | | From: ·Steve· | Sent: 1/19/2006 8:05 AM |
I thought I'd look around for more on this reaction, and I see both products given. Here's a page that gives MgO and NH3 as the products: http://dl.clackamas.cc.or.us/ch104-04/fourth.htm. Here's one that says Mg(OH)2 and NH3 are the products: http://en.wikipedia.org/wiki/Ammonia. Possibly MgO forms during the course of the reaction if both protons of one water molecule are transferred to the nitride ion. Leaving off the Mg2+ ion, we would have: H2O + N3-  NH2- + OH-
OH- + NH2-  NH2- + O2- This gives oxide ion, O2-, in the form of MgO, plus amide ion, NH2-, also with Mg2+ as its counterion. Then another water molecule could react with the amide ion to form ammonia: H2O + NH2-  NH3 + OH-
But I doubt if the reaction could be controlled to give the oxide, MgO, as the only product. Even if MgO forms, it would react with another water molecule, if available, to give Mg(OH)2. Possibly MgO results as long as excess water is not added. But for sure, in the actual experiment, after you add the water (5 to 10 drops, which is in excess with respect to the very small amount of Mg3N2 present), you have to reheat in order to decompose the Mg(OH)2. Interesting question!
>> I think they need to attend your class << Hey thanks... Bring 'em on!
Steve
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