How do I complete and balance an equation for reaction of NaOH with HCl?
544064
The two reagents are in aqueous form, which means that they exist in ions. Upon reaction, the H+ & OH- will form water, H2O & the remaining ions will remain aqueous in the solution;\n\nNaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)\n\nNo balance required...
955158
Find the moles of the acid CH3COOH and of the NaOH:\n\nn = C*V = 1*0.05 = 0.05 mol CH3COOH\n\nn = C*V = 1*0.05 = 0.05 mol NaOH\n\nAccording to the balanced chem. equation:\n\nCH3COOH + NaOH  --> CH3COONa + H2O\n\n1 mol of CH3COOH reacts with 1 mol of NaOH and gives 1 mol of H2O\n\nSo 0.05 mol CH3COOH reacts with 0.05 mol NaOH and gives 0.05 mol of H2O\n\nThat's the answer: 0.05 mol H2O
354896
1400 mL of NaOH needed...\n\nneed to make sure the concentrations are the same in order for them to completely neutralize\n\nM = Molarity... number of moles of solute in one liter of solution\n\nif you have 3 M for the sodium hydroxide you would need twice that amount in order to neutralize the 6 M of hydrochloric acid\n\nyou would need 1400 mL since the number of moles of the sodium hydroxide is exactly half of that of the hydrochloric acid.  With knowing that the number of moles NaOH is half, you would need double of what you have for the HCL...  700x2 is 1400 mL
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.
745017
Your question isn't really complete - the NaOH needs someting to react with.\n\nFor example:\n\nNaOH is used with lipids to make soap.\n\nNaOH + HCl -> H2O + NaCl
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
776525
Since you are pretty, here's how:\n\n1. Compute for the number of moles NaOH needed to prepare the solution.\nmole NaOH: 1 L * 0.5 mol/L = 0.5 mol\n\n2. Get the weight NaOH you need by multiplying the number of moles to the molecular mass of NaOH.\nweight NaOH: 0.5 mol * 40g/mol = 20g\n\n3. To prepare, weigh 20 grams of NaOH pellets. Dilute this with water to 1000 mL in a volumetric flask.
725692
To neutralise 2 moles of NaOH you need 1 mole of H2SO4\nTo neutralise 2 moles of NaOH you need 2 moles of HCl\nUsing the formula: moles = MV/1000 ;M = molarity; V = volume in dm3, then:\n\na) HCl would require twice the volume as H2SO4 to neutralise an 2 moles of NaOH.
903371
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
458346
2NaOH + CO2 = Na2CO3 + H2O
606155
NaOH + HCl ---> NaCl + H20
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
175516
I not sure what style you learned to use to balance chemical equation but here is one way to do it. \nWe have four elements to balance (Al,H,Cl,O)\nLeave the compounds with aluminum with one of each. (Al done)\nThe simplest compound Al is AlCl3. To match the Cl on both sides put a 3 in front on HCl (Cl done)\nIf you add up the H (hydrogen) on the left side of the equation you'll see 6 of them. Add a 3 in front of the H20 to balance that. (H done)\nDo a quick check and you'll see there are 3 oxygen on each side (0 oxygen)\nAll the elements have been balanced, yielding a final equation of:\n3HCl + Al(OH)3 ---> 3H20 + AlCl3