How would you balance H2SO4?
438197
This is sulfuric acid and is just as dangerous as hydrochloric acid. However, since you only have one reactant and there is no reaction, you do not have a chemical equation.
866732
do you want to know how to balance chemical equations? It's actually quite simple when you know how\n\nrather than asking for direct answers, it's better if you know how to do it for yourself\n\nI'm not going to answer your questions but show you how to do it for future reference\n\nH2 + O2-->H2O\n\nthe above equation is not balanced because there is only one oxygen atom on the right and two on the left\n\nnow, how do we balance this?\n\nfirst thing we need to do is ensure that there are equal amounts of atoms on the left and on the right\n\nso, to balance that equation\n\nwe will write\n\n2H2+O2-->2H2O\n\nthe equation is now balanced with respect to equal atoms on the left and right\n\nin words\n\n2  molecules of hydrogen  react with 2 oxygen atoms to produce 2 molecules of water\n\nthere are now 4 hydrogen atoms and two oxygen atoms on each side of the equation\n\nthe 2 in front of each compound represents the molecules and the subscripts represent the number of atoms
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
694411
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
506202
Write out the balanced reaction which occurs.\n\nSince both NaOH and H2SO4 are strong electrolytes (strong base, strong acid), we can assume both dissassociate completely and the reaction produced Sodium Sulfate and water,\n\n2NaOH + H2SO4 --> 2H2O + Na2SO4\n\nFind limiting reactant in order to find which  will be left over (either NaOH or H2SO4).\nIn order to do this, you need to know both the concentrations of the NaOH and the H2SO4 in order to find the initial number of moles of each.\n\nOnce you know how many moles of either OH- or H+ are left over, you can then find the pH, or pOH like this,\n\npH = -log ([H+])\npOH = -log ([OH-])\n\npH + pOH = 14\n\n*Remember to take into account that the final volume of solution is the sum of the intial volumes of each othe reactant solutions when calculating pH.\n\nEDIT:\nYou know, it just occured to me, I dont know why I even brought up pH, that was not what the question was asking.  but everything I said still stands in order to calculate the molarity...you just dont have to go the one extra step and calculate the pH by taking the negative log of the molarity concentration of H+ ions.\n\nAlso, for all of you out there who are just given answers with not other explanation, it would be nice if you stated what the concentration was of....salt? NaOH?  H2SO4?\nBecause of this ommision, I would highly inclined to disreguard their answers as numerically incorrect as well.
127716
based on the given balances equation.\n\nthe volumes could be:\n  \nSTP = 22.4L per mole\n\n 14.2L SO2 ( 1 mol H2S / 2 mol SO2) = 7.01 L of H2S\n\n and\n \n14.2L SO2 (3 mol O2 / 2 mol SO2 )  =  21.3 L of O2
1103908
you mean into a chloride?\neasy just add concentrated hydrochloric acid.the chloride will precipitate. and the sulfate will come out as hydrogen sulfide gas and water. maybe you need hydrogen to be passed in the mixture.\nna 2(so4)   + hcl  =      nacl   + h2s  + h2o
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
851560
solution...Bicarbonate and sodium. :)
1129845
You are right. The rest of the options are ionic where they will ionize into the cation and the SO42- ion but in the case of NaHSO4, the solution will first ionize into Na+ and HSO4- and then into H+ and SO42- i.e. there will still be some ions trhat remain HSO4- ions so it is the right answer
114804
No need to break it down to "molecular" other wise, try this: Sodium thiosulphate. Dont just use it coz they all vary in their Ph U know the basisity or acidity they concentrate.
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
788768
Just the usual stuff.  Use an electrolyte (H2SO4/HNO3), a couple of electrodes, a reduction table of half cell potentials, and some electricity.
1123465
Na2S + 2 HCl > H2S + 2 NaCl  \n\nH2SO4 + BaOH > H20 + BaSO4  \n\nNH4Cl + NaOH > NaCl + NH4OH\n\nAgNO3 + NaCl > AgCl + NaNO3\n\nnot sure about complete or reversible..:)
550368
I think you have a notational error: did you mean Al2(SO4)3 for aluminum sulfate?\nIf that is the case, then proceed as follows: \nAl(OH)3 + H2SO4 -> Al2(SO4)3 +H2O\n1. First, balance the Aluminum atoms:\n\n2Al(OH)3 + H2SO4 -> Al2(SO4)3+H2O (2 Al)\n2. Next, balance the Sulfur:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+H2O (3 S)\n3. Next, balance Oxygen:\n2AL(OH)3 + 3H2SO4 -> Al2(SO4)3+6H2O (18 O)\n4. Hydrogen is already balanced (12 H)\nGenerally speaking, the strategy is to balance the least represented element first until you move to the most represented.(here it was aluminum, followed by sulfur). Hydrogen is usually the last one you balance.
1029447
sulfuric acid is h2so4. convert 1600 kg to g so times by 1000. then take that number in grams and times by the molar mass of of sulfur (gSx1mol S/gS) then convert to moles of H2SO4 by using your balanced chemical equation(coefficients), with moles of S on bottom, moles of H2SO4 on top. everything cancels and your left with moles of sulfuric acid that you could make. if you need to take it another step to grams then you would then just use the molar mass of sulfuric acid.