How many grams of MgO(s) are produced by the complete reaction of 2.15 g of magnesium metal?
456620
Exactly 3.5653g of MgO will be made. Ignore the other guys' answers; they're retards.
106494
First, distill off the alcohol. With ethanol, the best you can get is 95% purity, the remaining 5% is water. There are cheap zeolites and hygroscopic salts such as CaCl2 or NaOH available which can further reduce the amount of water. If you are still not happy, you can get absolute alcohols by using magnesium. And magnesium ribbons or turnings are relatively cheap, too. Place (50-100 g) magnesium in a flask, put some amount of alcohol just to cover the magnesium. Leave for a few hours until you see white magnesium alkoxide being formed on the surface of the metal. Caution: exothermic reaction. Use the tallest distillation column you can find. Do not stir the mixture. Just leave it there. When you have now a lot of magnesium alkoxide you can add more alcohol (up to 2 liters). The alkoxide will destroy any water in the alcohol and you now can distill off your absolute alcohols. For methanol and ethanol, formation of the alkoxide takes only a few hours. For other alcohols, you might need to activate the magnesium by adding some solid iodine and heating to activate the magnesium. Don't distill to dryness. The remaining alkoxide reacts violently with water. Use alcohol-water mixture when disposing them. Sounds fun, doesn't it? =)
51598
2.73 grams. \n\nSame method as the others you've posted.
867158
2Mg+O2-->2MgO\n2MgO+2H2o---->2Mg(OH)2
977270
first you need a balanced equation.\n\n2Mg+O2===>2MgO\n\nNext convert everything given to mols.\n\n10.000 g Mg X 1 mol Mg / 24.30 g Mg (NOTE: this number is from the periodic table of the Elements) = 0.4115 mol Mg\n\n5.267 g O2 X 1 mol O2/ 32.00 g O2 (this is the molar mass of O2)= 0.1646 mol O2\n\nAt this point pick one of the mols you calculated.\n\n0.1646 mol O2 X 2 mol Mg (from balanced equation) / 1 mol O2 (from balanced equation)(for every 2 mol of Mg there is 1 mol of O2)= 0.3292 mol Mg needed to react with all of O2\n\nSince more Mg is available then needed (0.4115 mol > 0.3292 mol) this implies that O2 is the limiting factor. \n\nUse the limiting factor (O2) to get the mass of the MgO\n\n0.1646 mol O2 X 2 mol MgO / 1 mol O2= 0.3292 mol MgO\n\n0.3292 mol MgO X 40.30 g MgO/ 1 mol MgO = 13.27 g MgO\n\nessentially for all the other ones you are doing the same thing as before but with different numbers.
517398
Magnesium and trace amonts of mercury.
1022778
Mg2+ + 2HCl --> MgCl2 + H2 (gas)\n\nOkay, hydrogen.  As you can see in the above equation, for one atom of magnesium you'll produce two atoms of hydrogen.   \n\nLet's work out how much hydrogen you'll get in grams.\nAnd let's say that you have 1 gram of Mg.  First, what you need to do is work out how many atoms are in 1 gram of Mg.  \n\nThis is easy.  Simply:\n\n1 / atomic mass x Avagadro's constant = number of atoms.\n\nFor Mg, this equals:\n\n1 / 24.31 x 6.0221415×10^23 = 0.2477228 x 10^23 atoms.\n\nWe know that for every atom of Mg you'll produce 2 atoms of hydrogen.  So, that means you'll produce:\n\n2 x 0.2477228 x 10^23 = 0.4954456 x 10^23 atoms of hydrogen.\n\nTo work out how much 0.4954456 x 10^23 atoms of hydrogen weighs you just calculate, like this:\n\n0.4954456 / 6.0221415 x 1.008 = 0.0829 grams.\n\nAnd that's it.  1 gram of Mg will produce 0.0829 grams of hydrogen.
167906
atomic weight is 24.305g/gmol\natomic number is 12\n's right on the periodic table under Mg
712912
BaI2(aq) + MgSO4(aq)  --> BaSO4(s) + MgI2(aq)\n\nSo BaSO4 will precipitate.
790092
24g = 4g magnesium + 20g other elements\n1440/24 = 60 so need 60 times the amount\n4g * 60 = 240g magnesium\n20g * 60 = 1200g other elements\n240g + 1200g = 1440g\n\nSo 1200g of other elements
295894
magnesium sulfate heptahydrate or MgSO4 . 7H2O
895580
Mercury(II) oxide decomposes into its elements when heated.\nThere is no way that you would know this unless you had been taught about thermal decomposition. So all you need to worry about is the correct formula of oxygen gas.\n\nTherefore 2HgO --> 2Hg + O2.
924696
This isn't the final answer, but it is a clue:\n\nhttp://www.thesheetmetalshop.com/pdf_manual/Problem1.pdf
434401
I am pretty sure i did this right. using the CaCO3(s)as the limestone, i got 1178.934669 grams of CaO. (I didn't round the final answer, and w/in the problem itself, i rounded the atomic masses of Calcium (to 40.1), oxygen (to 15.9), and carbon (to 12)
900081
% yield is equal to the actual yield divided by the theoretical yield.\n\nThe balanced equation of the reaction is:\nBaI2 + MgSO4 ---> BaSO4 + MgI2\n\nFrom the information in the problem, we have:\n2.150(0.0600) = 0.129 moles of BaI2\nand\n1.980(0.0501) = 0.0992 moles of MgSO4.\n\nTherefore, your limiting reagent is MgSO4.\n\nBecause MgSO4 and BaSO4 have the same coefficients in the balanced equation, 0.0992 moles of BaSO4 should be produced by the reaction.\n\nYour theoretical yield can be calculated by:\n(molar mass of BaSO4: 233.37 g)\n0.0992 mol BaSO4 x (233.37 g/ 1 mol BaSO4) = 23.2 g\n\nSo, % yield = 18.862/23.2 x 100% = 81.3%\n\n(Note that sig figs will have a slight impact on calculations)\n\nHope that helped.
150925
Reaction One (CO is Co I'm assuming ... they are NOT equivalent) is a double replacement. The products will be aqeuous KNO3 and insoluble Co(OH)2\n\nReaction 2 is a redox. The product will probably be Mg2N3 (Magnesium Nitride) but I'm not entirely sure if that chemical formula is correct. The magnesium is oxidized and the nitrogen is reduced.\n\nAg + H2SO4 ---> Maybe Ag2SO4 as a redox.\n\nF2 + KBr ---> KF + BR2 Single Replacement
387221
B2O3(s)+Mg(s) yields MgO(s)+B(s)\nNotes: yields should be substituted with an arrow that is read as yields and in diboron trioxide the numbers are subscripts not superscripts.