How many grams of   Li2CO3   are produced from the complete reaction of   176.04 g   of   CO2 ?
952592
First thing you always need to do is balance your equation. While a balanced equation in the case doesn't matter math wise (because the mole ratio between CO2 and Li2CO3 is 1:1), it's still a good idea to get into the habit of balancing first. So:\n\n2LiOH + CO2 --> Li2CO3 + H2O\n\nUse this balanced equation, molecular weights of compounds, and molar ratio to find your answer:\n\n176.04 g CO2  x ( 1 mol CO2 / 44 g CO2) x (1 mol Li2CO3 / 1 mol CO2) x (73.88 g Li2CO3 / 1 mole Li2CO3) = 295.59 grams\n\nThen you use the information you just got (AKA Theoretical yield) to find the actual yield:\n\nPercent Yield =  (Actual Yield / Theoretical Yield)  x 100\n\nSo:\n\n (X / 295.59) = .9192   <--- this is the percent in decimal form so we don't have to worry about multiplying by 100. Just rearrange this equation to solve for X and you get your actual yield.\n\n X = (295.59)(0.9192)   =  271.71 grams\n\nHope this helped! Good luck with chemistry!!\n\n**EDIT** To Peter's response: He is correct. It should only be 4 sig figs, making it 271.7 grams. Sorry about my mistake :)
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
87793
44.00995 grams per mole
766197
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
757960
First you have to balence the equation.\n\n2LiOH + CO2 → Li2CO3 + H2O\n\n119.75 g LiOH x 1 mol LiOH     x    1 mol H2O  x    18.015 g H2O \n````````````````````` ----------------          --------------        ---------------\n``````````````````````23.948 g LiOH      2 mol LiOH        1 mol H2O\n= 45.04 g H2O (this is the theoretical yeild)\n\nto get % yeild you take auctual yeild (what you're looking for) and divide it up theoretical yeild (the above answer), then to turn it into a % you multiply it by 100. So % yeild= (auctual/theoretical)x100. \n\nsimple algebra gets auctual by itself. (%yeild x theoretical)/100\n\n91.72 x 45.04\n------------------ = 41.31 g H2O\n      100
911009
C6H6O + 7O2 yields 3H2O + 6CO2
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
519052
Na2CO3 + CaCl2 -> __2NaCl + CaCO3
748691
CH3COOH, remember the CO2 comes from the sodium carbonate so your question is poorly worded.
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
897570
Dimensional analysis is easier than it first seems.\n\nFirst you need to know the molecular weight of CO2.\n\nCarbon +2(Oxygen) = 12 + 2(16) = 44 grams per mole\n\n[The above step is not dimensional analysis.]\n\nLet's say somebody asks you how many moles 440 grams of CO2 is.\n\nThis is where we do dimensional analysis.\n\n(1 mole CO2 / 44 grams CO2) * (440 grams CO2 / ? CO2)\n\nThe / sign indicates "divided by", so imagine the parts over each other.  That makes it easier but I cannot do it here in text.\n\nThe first part of the above equation is stating that for every mole of CO2, there are 44 grams of CO2.  The second part says that we have 440 grams of CO2, and want to find out how many moles we have.\n\nThe trick here is to have the terms, one above the other, so that the units cancel out and you end up with the desired unit.  Our desired unit here is the mole - we want to know how many moles are in 440 grams of CO2.  So, that is why in the above, (1 mole CO2) is on top of (44 grams CO2).  If we allow 440 grams to be on top in the next term, then grams cancels out, and moles is the only unit left.  This is what we want.\n\nSo if we solve the above\n\n(1/44)*(440/1) = 440/44 = 10 moles of CO2.\n\nSo, 440 grams of CO2 is the same as 10 moles of CO2.\n\nhope this helps\n\nyay
322620
This is simple, since O=C=O (Carbon dioxide) includes the molecule C==O (carbon monoxide) and 28 gms of CO equals 1 mole, you'd need 1 mole of CO2 to produce 1 mole of CO.\n\nNow, if you want to take this to the next level, you'd need 44 gms of CO2 to equal one mole.
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
839352
If tha atomic mass of M is A, then the molecular weight is MW=2*A+60\n\nThe concentration of the M2CO3 solution is C=mole/V and since mole=mass/MW, C=mass/(MW*V)\n\nThen you do a titration for which we assume complete reaction and not reaching only the first end-point, so the reaction is\n\nM2CO3 + 2HCl -> 2MCl + CO2 + H2O\n\nTherefore 2*mole M2CO3=mole HCl =>\n2*C1*V1= C2*V2=>\n2*V1*mass/(MW*V) = C2*V2 =>\nMW= 2*V1*mass/(C2*V2*V)=\n=2*0.025*7.46/(0.1*0.027*1)\n=138\n\nand MW=2*A+60=138=> 2A=78 => A=39\n\nSo M, is potassium (K)
911120
2BF3 + 3Li2SO3 → B2(SO3)3 + 6LiF\n3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3H2CO3\n8Ag2S → 16Ag+S8
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.