HELP! How many moles of 2.0 molar HCl do i need to dissolve 1.5 cm of Mg?
487714
assuming the equation\n\n2HCl + Mg --> Mg2+ + H2 +2Cl-\n\ntake the moles of Mg you have... multiply by 2 and that is how many Moles of HCl you need. multiply by 2 because one Mg reacts with 2 HCl's so you need twice as much HCl. now that you have moles of HCl convert to volume of HCl, i.e. divide moles / concentration (in this case 2.O M) and that will give you liters of HCl needed.\n\n(.000382 moles *2)/2 = .000382 L = .382mL
973272
ok, first go to moles.\n\n24.0g LiOH (1mol/23.948g) = 1.0022mol\n\nnow get liters using molarity\n\n1.0022mol (1L/0.153mol) = 6.55L or 6,550mL \n\nRight, there ya go, remember that you have 3 sig.figs. so you may want to represent significance by using scientific notation. So...\n6.55x10^3mL\n\nIf you're ever stuck, just simply go to moles and that will get you started (and usually gets you some points on tests).
30238
No, HCL has nothing to do with the number of moles of CaCO3 that you used.  So you asume you used enough HCl so that all of your CaCO3 is dissolved.  All you need to do is devide 0.5215 g  with the molar weight of CaCO3 and you'll get your number of moles.\n\nHope this helps,\nUltimeaciax
931815
Your KCl example is a good one.  A 0.5 molar would have 1/2 mole/liter or 37.75g KCl in a liter. A 5% w/v KCl would be 50 g/liter; so the 5% is more concentrated.\n\nThe 2.5% w/v sodium bisulfate is 25g/liter and the .025M is 3 g/liter.\n\nIt all depends on the molecular mass.
739482
From chemical equation, moles of HCl reacted : moles of Cl2 produced = 4 : 1\nHence if 6.40 moles of HCl are used, moles of Cl2 produced = 6.40/4 = 1.60\nMass of 1.60 moles of Cl2\n= moles x Mr\n= 1.60 x (35.5x2)\n= 113.6 g\n\nHope this helps :)
977807
M = moles/L\nso, 0.220M = 0.220 moles per liter and there are 2 liters so there are...\n0.440 moles of Mg(NO3)2\n\nBUT... notice that there are 2 ions of NO3 for each molecule, so double it again and get 0.880 moles of NO3
609634
This question cant be answered without knowing the molarity of the HCl. I seriously doubt you were using 12M as another poster suggested. \n\nAlso. if all the Mg reacted the moles of Mg reacted would simply be 0.15/24.3 = 0.006. (Did you really use 0.15g? Thats a small amount) \n\nIf you didnt actually preform the rxn and/or dont know if all the Mg reacted you need to use the moles of Mg calculated above, and compare it to the moles of HCl used. M = moles / liters. So plug in the molarity of HCl used, and 0.050 L (converted 50mL) and solve for moles of HCl. Next use the mole ratio to determine the limiting reagent. If you dont know how to do that, your screwed.
318573
The reaction between Hydrochloric acid and Sodium Hydroxide is as follows,\n\nHCl + NaOH --> NaCl + H2O\n\nYou can see that there is a 1:1 ration between the number of moles of HCl needed to completely reacted with NaOH.\n\nSince you are given both a volume and concentration of the NaOH solution, you should be easily able to calculate the number of moles of NaOH availible.\n# mole = volume * concentration\n# moles = .1 Liters * .25 moles/L = .025 moles of NaOH\n\nIf you know you have .025 moles of NaOH, you know you need .025 moles of HCl to react it completely.\n\nIf you know that you have a .15 mole/Liter solution of HCl, you can calculate the volume required to get .025 moles lie this,\nLiters = moles / oncentration\nLiters = .025 moles / .15 moles/Liter = .01667 Liters or about 167 mL of solution.\n\n\nNext time do your own homework.
739325
According to Sigma-Aldrich the solubility of MgCl2.6H2O is 1M at 20C (or 20.3030 g/100ml solution), so it shouldn't be possible to prepare 4.5 M MgCl2 in just water. If there are more solutes involved which might increase the solubility of MgCl2 try dissolving those first and then the MgCl2.\n\nYet I saw some papers using 4.5 or 2.5 M...\n\nI don't know why there is this discrepancy between sigma-aldrich and the solubility provided in the link in the above answer which would correspond to 8 M if you consider 100ml of solution and not water. In reality you would have a significant change in volume when you add so much salt and the solubility would be much less but I don't know exactly how much. If you consider 167g of MgCl2.6H2O then you are adding 88.8 g of water which means that the total volume is at least 189 ml and thus the maximum concentration is 4.3 M.\n\nWikipedia gives a different number; 54.2g/100ml at 20C, without specifying if it is for the anhydrous or hexahydrate... so it would be 5.6 or 2.6M respectively... \n\nWhat a mess !!!\n\nMaybe you should also post the question in biology, since someone who has been doing such purifications can clarify the issue.\n\nAnother idea would be to prepare a 100ml saturated solution according to the lowest reported solubility (that of sigma-aldrich) using a 250 ml volumetric cylinder, and then from a pre-weighed amount of MgCl2.6H2O keep adding powder slowly until it doesn't dissolve any more. Weigh the powder that is left in order to find how much you dissolved, measure the final volume and calculate the molarity. If you achieve higher than 4.5 M, dilute.\n\nOR\n\nprepare a 4 M if the amount of powder that you couldn't dissolve is very small.\n\nor \n\nprepare 450ml of 1M and evaporate to 100ml of solution while praying that there will be no precipitation...\nThat's very desperate but I am running out of ideas...\n\nI must say the whole story is very fishy (finding very different solubility values)\n\nI'd go mad too...
961444
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
885670
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
844701
1) Calculate number of moles of NaOH:\n8.75 g/ (22.99+16.00+1.01) = 4.571 mol\n\n2) Calculate molarity in mol /liter\n4.751 mol / 25.4 mL = 0.187 mol/mL = 1.87 x 10^-4 mol/L
728109
Number of moles = mass/formula mass for a pure acid.\n\nBut acids in solution have a different equation:\n\nmoles = volume x molarity/1000\n\nThe volume in the above equation is in cm3.
961959
0,220M/L*2,20L= 0,484 moles of HCl\n12,00M = 0,484 moles/?\n? = 0,0403 L or 40,3 mL of 12,00 M HCl solution + 2,16 L of pure water, so you can get you 2,20 L of 0,220 M HCl solution
747941
To provide 3.0 moles of KOH we need 2 L of a 1.5M KOH solution
527028
This time i ans ur question but next time plz do ur homewrk ur self.\n\n1.\nWeight of NaOH = (4*.05) = .2 grams\nMoles of HCl =  Moles of NaOH = (.2/40) = .005\nVol. of HCl = .005*22.4 = .112 l = 112 ml\n\n2.\nWell let the grams be x\n1000x = 74.5*2.25*(75+x)\n(1000-167.25)x = 167.25*75\nx = 14.98 grams
306721
weight/M.W. = # of moles = 50 x 10E-3 in one litre\n\nso dissolve 6.057 g of Tris in 1 L of water and it should be 50mM