what is the balanced equation of aluminum sulfate------>aluminum oxide+sulfer trioxide?
927706
Al2(SO4)3 ==> Al2O3 + 3SO3
625685
Aluminium has a valanly of 3 Oxgygen has 2\nSo\n4Al + 3O2   -> 2Al2O3
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
580085
I think you are using Aluminum Sulfate, which is Al2(SO4)3. Molarity is defined as Moles per Liter, so the first thing to do is convert 210g Al2(SO4)3 to Moles. There are approximately 342.3g in one mole of Al2(SO4)3, so, after converting, you have about .613 Moles of Aluminum Sulfate. Then you can simply divide .613 by 2.75, giving you a molarity of .223.
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.
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
51552
AgNO3 + NaCl -----> AgCl + NaNO3\n\ntotal ionic equation\n\nAg+  +  NO3-  +  Na+  +  Cl-\n\n                      ----->\n\n                            AgCl  +  Na+  +  NO3-\n\nnet ionic equation\n\n    Ag+  +  Cl-  ----->  AgCl
363162
When you ask about the formula that is a pretty broad question.  The only formula I know is the balanced compound formula which is Ga2(SO4)3, direct combination formula which is 2Ga + 3S04 --> Ga2(SO4)3, or the decomposition formula which is, Ga2(SO4)3 --> 2Ga + 3S04.  If this isn't what you wanted I'm sorry, if you want you can email me and be more specific.
890642
Al + Fe(NO3)3 ---> Al(NO3)3 + Fe.\n\n2K + 2H2O ---> 2KOH + H2\n\n2KOH + CuSO4 ---> Cu(OH)2 + K2SO4\n\n3NaOH + H3PO4 ---> 3H2O + Na3PO4\n\nBest?
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.
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
802376
YOUR ANSWER IS CORRECT!  Here's how i did it:\n\nYou can find the # mol of NaOH used:\n\nMolarity = (# mol)/(volume solution)\n0.4 = (# mol)/(.075 L)\n\nSolving, you get 0.03 mol of NaOH.\n\nUsing the balanced equation:\n\n3 NaOH + H3PO4 ==> Na3PO4 + 3 H2O \n\nYou see that 3 mol NaOH = 1 mol H3PO4.  Thus, if we have 0.03 mol NaOH, that means we have 0.01 mol H3PO4.\n\nNow we can solve for the molarity:\n\nMolarity = (0.01 mol H3PO4)/(0.05 L) = 0.2 M
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
711424
The Equation should be:\nNaCl + H2SO4 = Na2SO4 + HCl\nSo the coefficients are (2,1,1,2)
457637
Aluminum Nitrate, Al(NO3)3, since, in a neutralization you take off any acidic hydrogen's, and any hydroxyls, and you combine the remaining ions in a way that the product molecule has no net charge. So\n Al(OH)3=Al(+3) + 3OH(-) \n HNO3 = H(+) + NO3(-)\nThe H(+) ions and OH(-) ions combine to form water, and each Al(+3) ion combines with 3 NO3(-) ions to produce Al(NO3)3 \n\nFormic acid (CHOOH) and sodium hydroxide(NaOH) produce Sodium formate, CHOONa, for the same reason as above.\n\nNote: The other H in Formic Acid is non-acidic, and so is not replaced by Na
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)