Iron + copper(II)  chloride → Iron(II)  chloride + copper any copper chloride remains in the beaker or not?!
824286
That depends on how much iron is added.\nCopper will continue to replace iron as long as any iron is left.\nIf xs iron was present, no copper would be left in solution...you would wind up with copper solid, suspended in iron chloride  soln...with any xs iron solid still unreacted.\nBut if you started with less than one mol of iron for each mol of copper, then yes, some copper would remain in soln.
882018
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
723259
This lab involves putting iron (steel wool) into a solution of copper sulphate. The reaction produces metallic copper and iron sulphate solution. I've never been really sure if it's iron(II) sulphate or iron(III) sulphate as the product. I've used both equations with good results. \n\n   Fe   +  CuSO4   --->  Cu  +  FeSO4\n   2 Fe   +  3 CuSO4   --->  3 Cu  +  Fe2(SO4)3\n\n\nFor prelab, make up a saturated copper(II) sulphate solution\nTrust me. Otherwise, the kids will be filtering for hours.  \n\nHave containers available for appropriate disposal of reagents. And make sure they clean up all of the bits of steel wool from the lab benches. Those little bits will haunt you for weeks if they don't.\n\n\n\n\nMethod\n\n1. Obtain 100 mL of copper (II) sulphate solution in\n   a clean, dry beaker. \n\n2. Obtain about 10 g of steel wool. Record the exact mass.\n \n3. Place the steel wool into the copper(II) sulphate solution.  Stir with a glass rod. When the solution loses its blue colour, add more copper(II) sulphate solution.\n\n4. Continue stirring (and adding solution, if necessary)until the strands of steel wool are no longer apparent, and only particles of a reddish brown solid are visible.\n\n5. Record the mass of a piece of filter paper.\n   Place it into a funnel.\n \n6. Pour the mixture through filter paper in a funnel.\n   Rinse any residue in the beaker into the filter        paper      with distilled water. Rinse the solid in the filter paper with distilled water.\n\n7. Allow the solid and filter paper to dry, then find the combined mass.\n\n\nAnalysis\n\n1. If iron was the limiting reactant, predict the mass of copper which should have been formed. Show work.\n  \n2. Calculate the percentage difference between  the mass of copper predicted in question 1 and the actual amount of copper produced. Show all work.\n\n3. Discuss possible explanations for your results in question 2.\n\n\n\nGood luck !
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)
945429
Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
854997
CuS + H2SO4 ===> H2S + CuSO4
951188
4. How many grams of zinc chloride can be formed from the reaction of 7.96 grams of zinc with excess HCL?\n\nZn + 2HCl → ZnCl2 + H2\n\n1 mole Zn produces 1 mole ZnCl2\nTherefore 65.7g Zn produces (65.7 + 2 x 35.5)g ZnCl2\nie ~65.7g Zn produces ~136.7g ZnCl2\nSo 7.96g Zn produces ~7. 96 x 136.7/65.7g ZnCl2\nie ~ 16.6g ZnCl2\n\n5. How many kilograms of copper sulfide could be formed from the reaction of 1.80 mol of copper with excess sulfur.\n\nCu + S → CuS\n\nSop 1 mole Cu produces 1 mole CuS\nTherefore 1.8 mole Cu produces 1.8 mole CuS\n1 mol CuS weighs ~(63.5 + 32.1)g = 95.6g\nSo 1.8 mol CuS weighs ~ 1.8 x 95.6g ≈ 172.1g = 0.1721kg\n\n6. If Copper(one) sulfide is partially roasted in air (reaction with 02), copper(one) sulfite is formed. Subsequently upon heating, the copper sulfite thermally decomposes to copper (one) oxide and sulfur dioxide. Write balanced chemical equations for these 2 reactions\n\nCu2S + 3O2 → Cu2SO3\nCu2SO3 → Cu2O + SO2
985144
My guess is dilute hydrochloric acid, to make soluble FeCL3, and AlCl3.
670034
Ag+ will precipitate in HCL solution as AgCl, but as for Ag(NO3), i am not so sure.
829125
I dont understand what you mean by "correcting certain formulas", but it appears Fe bears +1 oxidation state, if oxygen is -2. However, Fe does not bear such oxidation state, normally; it can bear 0, +2, +3, +4 and +6. With oxygen, what you will get, most likely, are FeO and Fe2O3. So, I would expect the the compound was a mixture of iron metal and one/several of these oxides. It appears that your compound could be 1:1 mixture of iron and iron(II) oxide.\n\nIf the question states that the compound was pure, check your calculation again; if you have done nothing wrong, the compound could be iron(II) peroxide [iron bears +2, oxygen bears -1]; it's not stable [I'm not sure whether it exists or not], or the strange iron(I) oxide [again, never seen this compound before, not sure it actually exists or not].
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
638216
All of these may have an effect on absorbance since they all have the potential to affect the final concentration of the solutions being measured; In (a), you don'y know what might be in the wet stuff in the beaker, so your absorbance might go up.  In (b), because the pipet has not been "wetted" with your solution, you might pick up less than you thought, so your concentration might be lower and give a lower absorbance.  I'll let you figure out (c) for yourself since I can't be sure what you typed!
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
816320
Since chloride is an halogen, it has a charge of -1.\nBecause a molecule always have a charge of zero, the iron must be a positive charge.\nThere are three Cl, for a total of -3, and only one Fe, that has to have the whole +3 all to itself to balance out the -3 of the 3 Cl.