IF 1.5 g of sodium carbonate are mixed with 95. ml of 6 M hydrochloric acid, how many liters of carbon dioxide
600936
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O\n\nno. of mol of Na2CO3 reacted\n= m/M\n= 1.5/106.2\n= 0.01412 mol\n\nno. of mol of HCl reacted\n= 95/1000 x 6\n= 0.57 mol\n\nsince Na2CO3 is the limiting reagent, the no. of mol of CO2 liberated = 0.01412 mol\n\nMass of CO2 liberated\n= 0.01412 x 44.01\n= 0.6214g\n\nVol. of CO2 gas \n= 0.6214 x 1.96\n= 1.217\n= 1.22g (3s.f.)
980259
2(LiOH) + CO2 → Li2CO3 + H2O\n\nHow many grams of H2O you make will depend on how many grams of LiOH and CO2 you start with
321990
Baking Soda is also known by its chemical name of\nSodium Bicarbonate (NaHCO3).\n\nNaHCO3 will react with acid (for example, HCl) to produce a Soium Salt and Carbonic Acid.\nNaHCO3 + HCl --> NaCl + H2CO3\nThe Carbonic acid decomposes to form water and Carbon Dioxide gas,\nH2CO3 --> H2O + CO2\n\nSo the over all reaction is as follows,\nNaHCO3 + HCl --> NaCl + H2O + CO2\n\nAs you can see, there is a 1:1 ration of Sodium Bicarbonate to Hydrochlorid acid needed to neutralize the solution, so for every 1 mole of HCl you have, you need 1 mole of NaHCO3 to completely react it.\n\nIn 4 Liters f 6.25 Molar HCl solution, there are 25 moles of HCl, thus you need 25 moles of NaHCO3.\n\nThe molar mass of NaHCO3 is 84.01 g/mol.\nIf you need 25 moles, you would need 2100.25 grams of NaHCO3 or about 2.1 kg (4.62 pounds).
478026
1400 mL of NaOH needed...\n\nneed to make sure the number of moles are the same in order for them to completely neutralize\n\nM = Molarity... number of moles of solute in one liter of solution\n\nif you have 3 M for the sodium hydroxide you would need twice that amount in order to neutralize the 6 M of hydrochloric acid\n\nyou would need 1400 mL since the number of moles of the sodium hydroxide is exactly half of that of the hydrochloric acid. With knowing that the number of moles NaOH is half, you would need double of what you have for the HCL... 700x2 is 1400 mL\n\nthanks redsuns :D
920532
1) Calculate the number of moles in 95.80 gram of LiOH :\n1 mol of LiOH is 6.941+16.00 + 1.008 = 23.95 gram \nHere we have 95.80 / 23.95 = 4.00 mol\n2) Balance the reaction: \n2 LiOH + CO2 → Li2CO3 + H2O\nFrom the reaction equation follows that 2.00 mol of Li2CO3 were formed.\n3) Calculate the mass of Li2CO3:\n1 mol of Li is 2*6.941+12.01 + 3*16.00 = 73.89 gram \nHere we have 2*73.89 = 147.8 gram of Li2CO3
952447
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).
764912
If I understand correctly what you need then this is what you do:\n(by the way we will take He as 83.7 because it is over 100% other wise.)\nmultiply the molecular or atomic weight of each species by the fraction (percent divided by 100).\n\nMW(CO2) = 44\nMW(N2) = 28\nAW(He) = 4\n\n44*0.11 + 28*0.053 + 4*0.837 = 9.67 g/mol\n\nThis is the equivalent "molar mass" of the mixture.
541101
First you need the equation for the reaction of CO2 with phenyl magnesium bromide to get the mole ratio. (i get one to one.)\n\nSecond, calculate the number of millimoles of CO2 required to react.\n\nLastly, multiply the number of millimoles of CO2 by its molar mass. I get 88 mg.
984571
1 litre (1000 ml) of 1 M solution of NaCL contains 23 + 35.5 gms =58.5 gms (ie 1 molecular wt of sodium chloride in gms dissolved in 1 lit of solution)\n\nHence 1 lit of 0.360 M solution  will contain 0.360 X 58.5 gms or 21.06 NaCl (Sodium Chloride) \n\nTo find out exact vole in ml which contain 21.14 find volume by this formula 2.14 divide by 21.06 multiply 1000 , (ie 2.14/21.06 * 1000 ) 101.6 ml\n\nPS\nbasic assumption is 0.360 M indicates .360 Molar solution. \n(ie: M stands for Molar)
860818
From the reaction, 2 moles LiOH react with one mole CO2.\n\nSo find out how many moles of LiOH you have in 454 g LiOH.  Then divide by two (according to the first sentence above) and you have the number of moles of CO2.\n\nFinally multiply the number of moles of CO2 by the molecular weight of CO2 and you will get the mass of CO2.\n\nIf you don't understand you should do something else.  Nothing to do with chemistry.
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
119509
my question is:\n\nNaHCO3(s) + NaOH(aq) -> Na2CO3(aq) + H2O(l)\n\n3.25 moles of Na2CO3 is produces.
619470
o.k., in short, d < a < c < b. This is basic material, so I'll give a generously detailed explanation.\n\na) 2 moles N2 * (28 g N2/ mole N2) = 56g N2\n\nb) 1 mole KCl * [(39.1gK +35.5gCl)/ mole KCl] = 74.6 g KCl\n\nc) 9 x 10^23 molecules CO2 * (1 mole CO2/ 6.02 x 10^23)*([12gC + 16gO + 16gO]/ mole CO2) = 66g CO2\n\nd) 2 moles LiF * [(6.94g Li + 19gF)/ mole LiF] = 51.88 g LiF
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
575750
This can be done using the formula M1V1=M2V2.  For example, if you have 2M HCl and wanted 2L of .5M HCl, then putting that into this equation as 2(V1)=(.5)(2).  Solve accordingly.  This is just an example.  This formula can be used in a variety of ways that all.  But, to make it to the desired you, a volumetric flask is needed in which water is added until the etched line on the flask to equal the preset volume of the flask.
921837
NaHCO3 --> Na+ + HCO3-\nCa(NO3)2 --> Ca(2+) + 2NO3-\nAl2(CO3)3 --> 2Al(+3) + 3CO3 (-2)
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