What is the balanced equation of copper(ll) sulfate pentahydrate reacting with iron?
83000
The net ionic equation is:\n\nFe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)
116069
The balance is :\n\n     H2O2(aq) -> O2(g) + 2H+(ag) + 2e-\n MnO2(s) + 4H+(aq)  +  2e-  ->  Mn2+(aq)  + 2H2O\n \n Combine the two half-reactions and cancel on both sides yield the     complete balance equation.\n\n MnO2(s) + H2O2(aq) + 2H+(aq) -> Mn2+(aq) + O2(g) + 2H2O\n\n this is an oxidation-reduction reaction\ntype  decomposition reaction.
812692
Ferric sulfite, or iron (III) sulfite
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
903862
Sorry Voyager. Copper is not so reactive a metal that it will react with sulphuric acid.\nYou need either solid copper (II) carbonate or solid copper (II) oxide for the salt.\nSince it is a sulphate, sulphuric acid is used together with the solid.\nHeat the acid so that the reaction can proceed at a faster rate. Add solid till no more will react (no bubbles of gas produced, and the solid remains insoluble). This is to ensure that all the acid is used up. Filter to remove unreacted solid and heat the filtrate, which is the copper (II) sulphate solution, till saturated (approximately 1/3 to half the original volume left would be a good gauge). Let the hot saturated copper (II) sulphate solution cool and after some days, when the excess water evaporates off, you should get hydrated copper (II) sulphate crystals.\nDo not heat till dryness as this will evaporate off the water of crystallization needed for the blue copper (II) sulphate crystals to form. Without the water of crystallization, you would get anhydrous copper (II) sulphate powder (white) and not the blue crystals.\n\nHope this helps :)
819316
CuSO4+Fe = FeSO4+Cu\nCompute the number of atoms of CuSO4 in 7 gms of it, the number of atoms in 2gm of Fe. Choose the smaller number of the two. That is the number of atoms of Cu formed. Compute its weight.
267625
Iron (III) Chloride Hexahydrate
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
256406
4Fe3+ + 9O2 ->  6Fe2O3
908510
It should be Fe2+ + 2OH- --> Fe(OH)2 since you not only need to balance the elements but you also need to balance the charges.
946990
FeCl3 + NaCl -> FeCl2 + NaCl3\n\n\nFeCl3 + NaCl -> FeCl2 + NaCl3\n\nthese reactions are not possible \nas this reaction can take place if iron(fe) was more reactive than sodium or fe was less electropositive than na\nhence \nthese reactions are not possible
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
590452
(NH4)2 SO4  + (FeSO4)2 + 2H2O____(\nNH4)2SO4Fe2(SO4)3.24H2O\n\nAmmonium ferric sulphate
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
707894
Cu 2+ ClO3 1-\n\nCu(ClO3)2