for the reaction LiOH + CO2 --> Li2CO3 + H2O?
956231
Balance the equation\n2LiOH + CO2 --> Li2CO3 + H2O\n\nFind the molecular weight of CO2\n\n1 mol C * 12.01g/mol = 12.01g\n2 mol O * 16.00g/mol = 32.00g\n                                       44.01g/mol CO2\n\nconvert from grams CO2 to moles CO2\n\n88.02g CO2 * 1 mol/44.01g = 2 mol CO2\n\nconvert from moles CO2 to moles Li2CO3\n\n2 mol CO2 * (1 mol Li2CO3/1 mol CO2) = 2 mol Li2CO3\n\nconvert from mol Li2CO3 to grams Li2CO3\n\nfirst we need the molecular weight\n\n2 mol Li * 6.941g/mol = 13.88g\n1 mol C * 12.01g/mol = 12.01g\n3 mol O * 16.00g/mol = 48.00g\n                                       73.89g/mol Li2CO3\n\nNow we continue\n\n2 mol Li2CO3 * 73.89g/mol LiCO3 = 147.78g Li2CO3 produced\n\nHope this helps!
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
145111
The reaction formula is\n\nCH2COOH                           \nCH(OH)COOH (aq) + NaHCO3 (aq)----> \nCH2COOH                            \n\nCitric Acid\n\n       CH2COONa\n CH(OH)COONa    (aq)+ H20 (l) + CO2(g)\n       CH2COONa
332502
Yeah, I've actually done this reaction before several times. \n\nTheoretically, this is a metathesis reaction, also called a double-replacement reaction.  Essentially, the ions switch partners.  \n\nCaCO3 + 2HCl ---> CaCl2 + H2CO3\n\nNow, the CaCl2 is a soluble salt, so that's actually just like ions floating around in solution. \nThe H2CO3 is carbonic acid, which is actually just carbon dioxide, CO2, dissolved in water.  The carbon dioxide will immediately come out of solution and bubble out, which is totally observable, just like putting baking soda and vinegar together.  \n\nThe more complete reaction full ionic equation is the following:\nCaCO3 + 2H+ + 2Cl- ---> Ca2+ + 2Cl- + H2O + CO2
744031
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2\n\n0.1L * 0.2 mol/L = 0.02 mol HNO3\n\n0.02 mole HNO3 * 1 mol CO2/2 mol HNO3 = 0.01 mol CO2\n\nNow plug 0.01 mol CO2 into PV=nRT, using STP conditions.\n\nPV=nRT\n\nV = nRT/P\n\nV = (0.01)(.0821)(298)/(1)\n\nV = 0.24 L\n\n\n*** The person above me did it WRONG!  He confused sodium carbonate with sodium bicarbonate.  Also, he completely forgot to do the stoichiometric conversions.  He solved for moles of HNO3, and then plugged it into PV=nRT.  You must first convert moles of HNO3 to moles of CO2 with stoichiometric conversions, using the BALANCED equation.
122744
Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
868179
2Rb + 2 H20 --> 2RbOH + H2
50281
This is an acid base reaction, but the other answer is incorrect about the product. There is no Cl present. Here’s the balanced equation:\n\n2NaOH (s) + H2CO3 (aq) ➔ Na2CO3 + 2H2O\n\nNaOH is the base. H2CO3 is the acid.
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
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 :)
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
746063
CO3^2- + 2H^+ ===> CO2 + H2O
857898
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
873085
88.02 g of CO2 are equivalent to 88.02/44 moles of CO2 = 2 moles\nAccording to the equation 1 mol LiOH reacts with 1 mol CO2,\n\n2 moles CO2 will require 2 mol LiOH.\n\nTo find out the gram amount multiply 2 moles of LiOH by its molecular weight\n\n2 x 23.95 = 47.9 grams LiOH.