2.74 grams of barium reacted with 250cm3 of water. how many moles of barium reacted?
399392
The molar mass of Barium is 137.33 g/mol.\n\nDivide the mass of Barium reacted by the molar mass to obtain the number of moles of Barium reacted.\n\n2.74 grams / 137.33 g/mol = .01995 moles of Ba
949044
1) put them into the solution of bromine, the substance which makes the solution lose its colour is NaI\n2NaI + Br2 => 2NaBr + I2\n2) dissolve them in water, barium sulfate is the only one which cannot be dissolved in water\n3) Na2CO3 + 2HCl => 2NaCl + H2O + CO2\nNaHCO3 + HCl => NaCl + H2O + CO2
817888
The number of moles of KOH used is:\n\n3.75 g / 56.1 g/mol = 0.0668 moles\n\nin 250 mL (1/4 L) this is 0.0688 * 4 = 0.267 M\n\nM1 * V1 = M2 * V2  or  M2 = (M1 * V1)/V2\n\n= (0.267 M * 27mL)/100mL = 7.209 / 100\n\n= 0.0721 M
900081
% yield is equal to the actual yield divided by the theoretical yield.\n\nThe balanced equation of the reaction is:\nBaI2 + MgSO4 ---> BaSO4 + MgI2\n\nFrom the information in the problem, we have:\n2.150(0.0600) = 0.129 moles of BaI2\nand\n1.980(0.0501) = 0.0992 moles of MgSO4.\n\nTherefore, your limiting reagent is MgSO4.\n\nBecause MgSO4 and BaSO4 have the same coefficients in the balanced equation, 0.0992 moles of BaSO4 should be produced by the reaction.\n\nYour theoretical yield can be calculated by:\n(molar mass of BaSO4: 233.37 g)\n0.0992 mol BaSO4 x (233.37 g/ 1 mol BaSO4) = 23.2 g\n\nSo, % yield = 18.862/23.2 x 100% = 81.3%\n\n(Note that sig figs will have a slight impact on calculations)\n\nHope that helped.
558772
the correct answer is 1.30.  the answer would be two if the concentration was 0.01M.  you just take the negative log of the concentration of hydronium ion.
572470
a.  4H20 + 4NO2 + 02--------->4HNO3 + 2 H2O\n      Al(OH)3+ 3HCl------------> AlCl3+ 3H2O\nb. not sufficient as Al(OH)3 is 12/78 moles which is able to react with .462 moles of HCl and here we have .9444 moles of HCl\n\n9. it is done by equilliberium reaction\n and u written Na reaction with H2 to form hydride
124995
Barium chloride is soluble in water. Generally we dont say a molecule is reacting when it dissolves. (Like when NaCl dissolves in water.) \nWhen barium chloride dissolves the following happens\n\nBaCl2+2H2O --> Ba(OH)2+2HCl \n\nThe solution becomes acidic due to the formation of HCl. Though there is a chemical change, these type of changes are   not said chemical reaction. Rather they are dissolving of ions due to hydration.
890894
C12H22O11+12O2-------12CO2+11H2O+E
1153335
barium acetate and carbon dioxide and water
986414
Start by balancing the equation:\n2Ca3(PO4)2 (s) + 6H2SO4 (aq)-----> 6CaSO4 (s) + 4H3PO4(aq)\n\nThen figure out the molar masses of each of the reactants:\nCa3(PO4)2 = 3(40.078) + 2(30.973762) + 8(15.9994) = 310.176724\n\nH2SO4 = 2(1.00794) + 32.065 + 4(15.9994) = 98.07848\n\nDivide each of your known masses of reactants by its respective molar mass to determine how many moles of each you have, and divide this number by the coefficient (the number in front) of the reactant to determine how many "parts" of each you have:\n(1000/310.176724)/2 = 1.612 parts 2Ca3(PO4)2\n(980/98.07848)/6 = 1.665 parts 6H2SO4\n\nTake the smaller number (since once you run out of one reactant, the reaction is over) and figure out how many moles of calcium sulfate and phosphoric acid you'll produce:\n(1.612)(6)(40.078 + 32.065 + 4(15.9994)) = 1316.7518832g CaSO4\n\n(1.612)(4)(3(1.00794) + 30.973762 + 4(15.9994)) = 631.872933536g H3PO4\n\nSince the total mass of reactants and products needs to be the same, you can check your (er, my) work by totalling up both sides:\n(1.612)(2)(310.176724) + (1.612)(6)(98.07848) = 1948.625g\n1316.7518832 + 631.872933536 = 1948.625g
903475
Well, we first find the molar mass of SrS2O3.  I'm getting 199.76 grams/mole.  How many moles have we got?  2.30/199.76=.0115 moles.  We have 3.34-2.3=1.04 grams water.  Water is 18 grams/mole, so we have 1.04/18 = .0578 moles.  The mole ratio is .0115 to .0578, which is about 1 to 5 or .2 (with a small bit of error; you actually get .199).  So for every 1 part SrS2O3, you should get 5 parts water.
464799
I assume that you have an equation such as A and B react together to give C. Work out the molar ratios of each, i.e. is it 1 equivalent of A reacted with 1 equivalent of B to give 1 equivalent of C?\nAssuming it is a 1:1:1 ratio, then look at the amounts you actually reacted - chances are that you did not use 1 mole of each of the reactants. Work out the molar quantities of each reactant. (mass used/molecular weight of compound)\n\nIf you used 0.5 moles of each, then you would get 0.5moles of product and just convert this to a mass.\n\nIf one of the reactants is in excess, e.g. 0.5 moles of A and 0.8moles of B was used, ignore B because as there is only 0.5 moles of A you would only make a maximum of 0.5 moles of C.\n\nHope this helps
793636
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
892172
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
899377
Ba(OH)2 + 2CH3COOH ----> (CH3COO)2 Ba\n\nmole CH3COOH: 12g/ 60g/mol = 0.2 mol\nmole Ba(OH)2: 0.2 mol/ 2 = 0.1 mol\n - since 2 mol acetic acid is required for every mole Ba(OH)2\n\nIf concentration of Ba(OH)2 is 0.1M, volume required to react is 0.1 mol/ 0.1 M = 1L or 1000 cm3
715558
OK, but you need to look up atomic weights or calculate molar mass.\n\n   Grams = Moles x molar mass  OR  moles x atomic mass\n\n    Moles = Grams/(molar mass) OR Grams/(atomic mass)
11182
stochiometrically\nthe reaction is as under\n2H2 + O2 --> 2H2O\n\nhere 2 moles H2O is produced. Now to produce 2 moes of water 2 moles H2 is required. So to produce 8moles water 8 moles H2 will be required