how many grams of   Li2CO3   are produced from the complete reaction of   95.80 g   of   LiOH ?
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
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
118416
MY ANSWER IS:\n\n1 mol/g of LiCl = 35.4g
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
911009
C6H6O + 7O2 yields 3H2O + 6CO2
788070
Well its easy dear.\n\n2Li (s) + Cl2 (g) ----> 2 LiCl (s)
458346
2NaOH + CO2 = Na2CO3 + H2O
254672
You answer this question by formula weight ratios.  One mole of Na2CO3.10H2O has a mass of 286.14 g.  It also contains 13 moles of oxygen (3 in the carbonate + 10x1 each water), which is 13x15.9994 = 207.99 g.\n\nThe mass percent oxygen is then (amt in question / amt total)x100% = (207.99/286.14)x100% = 72.688%, or 72.7% to the nearest tenth.
907209
Na2CO3 + NaHCO3  \n\nmoles Na2CO3 = 0.070L * 5.5 mol/L = 0.385 mol\nmoles of Na+ from Na2CO3 = 0.77 mol\nmoles NaHCO3 = 0.030L * 1.0 mol/L = 0.030 mol\nmoles of Na+ from NaHCO3 = 0.03 mol\ntotal moles of Na+ in the solution = 0.80 mol\nsodium ion (Na+) concentration = 0.80 mol/0.1L = 8.0M
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
433273
Na2CO3 + 2HCl = CO2 + H20 + 2NaCl (aq)\n\nThis reaction is a double displacement reaction because oxygen from the sodium carbonate combines with the hydrogen in the hydrochloric acid to form water, and the sodium goes with chlorine to form sodium chloride, making carbon dioxide in the process.
844238
Li2O + H2O --> 2 LiOH\n\nIf you include the H2 it will never ballance.  You will not end up with H2 as a product anyway since OH will react with Li and the remaining H will react with the LiO hence 2 of LiOH.  Now on both sides of the equation you have 2 - Li, 2-O and 2-H.  If you did want to include a water molecule on the produst side then the equation would read as follcws:\n\nLi2O + 2H2O --> 2LiOH + H2O\n\nthis equation ballances with 2-Li, 3-O and 4-H on each side\n\nThe first equation shaow the ideal combination of the two compuonds (what most teachers are looking for) and the seccond equation shows a instance where there is not enough Li2OH to completely react with the all of the water
603847
if its 2% by mass, thats 2 grams of KCl for every 100 grams of solution its in. \n\n2g KCl= 2g / (74.55g/mol)KCl = 0.0268mol KCl\n\nBecause there is one mole of Cl in KCl, this is also the number of moles of Cl in 100 grams of solution.\n\n0.0268mol Cl * (35.45g/mol)Cl = 0.951gCl/100gsolution\n\nThis is 0.951%\n\nnote: 35.45g/mol is the molar mass of Cl and 74.55g/mol is the molar mass of KCl. These were obtained from the periodic table.
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.
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
770748
Oxygen is the limiting reactant....you would need 5.6 moles of it for a complete reaction.\n\nMoles of CO2...2.8.  It is 1/1 with methane\n\nGrams 2.8 X 44 = 123.2 grams
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2