H2So4 Kmno4 Feso4 = Fe2(So4)3 H2O Mnso4 K2So4
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Bạn đang xem: H2so4 kmno4 feso4 = fe2(so4)3 h2o mnso4 k2so4
Charchit Tailong
2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2OFeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2OFirst of all, let's concentrate on what is really important - on the net ionic reaction:Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2OAll other ions (K+and SO42-) are only spectators và we don't need them khổng lồ balance the equation. Besides, charges are what is really important in the half reactions method.At first glance you can see that in the reaction iron gets oxidized and permanganate gets reduced:Fe2+-> Fe3+MnO4--> Mn2+We will start equation balancing with balancing these half reactions - using electrons lớn balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. Lớn balance charge we will showroom one electron on the right side:Fe2+-> Fe3++ e-We can use electrons safely, as the final step of balancing will be electron cancellation.Now we have khổng lồ balance permanganate reduction half reaction:MnO4--> Mn2+Before balancing charges we have khổng lồ balance atoms. What to vày with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can địa chỉ cửa hàng H+on the left and H2O on the right:MnO4-+ H+-> Mn2++ H2OUsing simple balancing by inspection we will địa chỉ cửa hàng two coefficients khổng lồ balance atoms:MnO4-+8H+-> Mn2++4H2OOnce the atoms are balanced it is time lớn balance charge with electrons - there is +7 charge on the left side of the equation & +2 on the right side. To balance charges we have to địa chỉ cửa hàng 5 electrons on the left:MnO4-+8H++5e--> Mn2++4H2OAt the moment we have two balanced half reactions. Here comes the final trick - we địa chỉ these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to lớn use just numbers of electrons, although if often means you will have to find lowest coefficients later). To lớn have five electrons in both equations we have khổng lồ multiply first equation by 5:5Fe2+->5Fe3++5e-and when we địa chỉ cửa hàng both half reactions we will getMnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e-Canceling out the electrons:MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+And the reaction is balanced.Balancing hydrogen & oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can địa chỉ H+on the side where there is oxygen excess và water on the second, just as we did in the above example. But we can also use OH-and water to vì the trick, for example half reaction:ClO--> Cl-is not balanced, but once we add water & OH-:ClO-+ H2O -> Cl-+2OH-we have not only balanced atoms but we are also ready lớn balance charge by adding two electrons on the left:ClO-+ H2O +2e--> Cl-+2OH-and the half reaction is ready khổng lồ be used. General rule says that if the reaction takes place in acidic conditions we use water và H+to balance oxygens, & if the reaction takes place in basic conditions - we use OH-and water. Don't worry if it looks lượt thích the reaction produces H+in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.
Rishi Sharma
ccevents.vn Faculty 646 Points
2 years ago
Xem thêm: Thế Vận Hội Olympic Hiện Đại Được Tổ Chức Mấy Năm 1 Lần ? Sơ Lược Về Đại Hội Thể Thao Thế Giới (Thế Vận Hội
Dear Student,Please find below the solution khổng lồ your problem.2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2O FeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2O First of all, let's concentrate on what is really important - on the net ionic reaction: Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2O All other ions (K+and SO42-) are only spectators and we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method. At first glance you can see that in the reaction iron gets oxidized & permanganate gets reduced: Fe2+-> Fe3+ MnO4--> Mn2+ We will start equation balancing with balancing these half reactions - using electrons khổng lồ balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. Lớn balance charge we will địa chỉ one electron on the right side: Fe2+-> Fe3++ e- We can use electrons safely, as the final step of balancing will be electron cancellation. Now we have lớn balance permanganate reduction half reaction: MnO4--> Mn2+ Before balancing charges we have lớn balance atoms. What to vì with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can địa chỉ cửa hàng H+on the left và H2O on the right: MnO4-+ H+-> Mn2++ H2O Using simple balancing by inspection we will địa chỉ two coefficients to lớn balance atoms: MnO4-+8H+-> Mn2++4H2O Once the atoms are balanced it is time lớn balance charge with electrons - there is +7 charge on the left side of the equation & +2 on the right side. To balance charges we have to showroom 5 electrons on the left: MnO4-+8H++5e--> Mn2++4H2O At the moment we have two balanced half reactions. Here comes the final trick - we địa chỉ cửa hàng these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is lớn use just numbers of electrons, although if often means you will have lớn find lowest coefficients later). Lớn have five electrons in both equations we have lớn multiply first equation by 5: 5Fe2+->5Fe3++5e- & when we địa chỉ cửa hàng both half reactions we will get MnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e- Canceling out the electrons: MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+ and the reaction is balanced. Balancing hydrogen & oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. Lớn balance oxygen we can showroom H+on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH-and water to vì the trick, for example half reaction: ClO--> Cl- is not balanced, but once we showroom water and OH-: ClO-+ H2O -> Cl-+2OH- we have not only balanced atoms but we are also ready to balance charge by adding two electrons on the left: ClO-+ H2O +2e--> Cl-+2OH- và the half reaction is ready khổng lồ be used. General rule says that if the reaction takes place in acidic conditions we use water và H+to balance oxygens, và if the reaction takes place in basic conditions - we use OH-and water. Don't worry if it looks like the reaction produces H+in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.Thanks & Regards
Rishi Sharma
ccevents.vn Faculty 646 Points
2 years ago
Xem thêm: Combo Chăm Sóc Da Mặt Cho Nam Giới Tốt Nhất Hiện Nay, Mỹ PhẩM Cho Nam
Dear Student,Please find below the solution to your problem.2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2O FeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2O First of all, let's concentrate on what is really important - on the net ionic reaction: Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2O All other ions (K+and SO42-) are only spectators và we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method. At first glance you can see that in the reaction iron gets oxidized & permanganate gets reduced: Fe2+-> Fe3+ MnO4--> Mn2+ We will start equation balancing with balancing these half reactions - using electrons lớn balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. To lớn balance charge we will showroom one electron on the right side: Fe2+-> Fe3++ e- We can use electrons safely, as the final step of balancing will be electron cancellation. Now we have lớn balance permanganate reduction half reaction: MnO4--> Mn2+ Before balancing charges we have to lớn balance atoms. What to vày with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can địa chỉ H+on the left và H2O on the right: MnO4-+ H+-> Mn2++ H2O Using simple balancing by inspection we will địa chỉ cửa hàng two coefficients to balance atoms: MnO4-+8H+-> Mn2++4H2O Once the atoms are balanced it is time to balance charge with electrons - there is +7 charge on the left side of the equation & +2 on the right side. Khổng lồ balance charges we have to add 5 electrons on the left: MnO4-+8H++5e--> Mn2++4H2O At the moment we have two balanced half reactions. Here comes the final trick - we add these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to lớn use just numbers of electrons, although if often means you will have khổng lồ find lowest coefficients later). Lớn have five electrons in both equations we have lớn multiply first equation by 5: 5Fe2+->5Fe3++5e- and when we địa chỉ cửa hàng both half reactions we will get MnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e- Canceling out the electrons: MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+ & the reaction is balanced. Balancing hydrogen and oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can showroom H+on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH-and water to vì chưng the trick, for example half reaction: ClO--> Cl- is not balanced, but once we địa chỉ water and OH-: ClO-+ H2O -> Cl-+2OH- we have not only balanced atoms but we are also ready khổng lồ balance charge by adding two electrons on the left: ClO-+ H2O +2e--> Cl-+2OH- và the half reaction is ready khổng lồ be used. General rule says that if the reaction takes place in acidic conditions we use water and H+to balance oxygens, & if the reaction takes place in basic conditions - we use OH-and water. Don't worry if it looks lượt thích the reaction produces H+in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.Thanks and Regards
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Charchit Tailong
2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2OFeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2OFirst of all, let's concentrate on what is really important - on the net ionic reaction:Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2OAll other ions (K+and SO42-) are only spectators và we don't need them khổng lồ balance the equation. Besides, charges are what is really important in the half reactions method.At first glance you can see that in the reaction iron gets oxidized and permanganate gets reduced:Fe2+-> Fe3+MnO4--> Mn2+We will start equation balancing with balancing these half reactions - using electrons lớn balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. Lớn balance charge we will showroom one electron on the right side:Fe2+-> Fe3++ e-We can use electrons safely, as the final step of balancing will be electron cancellation.Now we have khổng lồ balance permanganate reduction half reaction:MnO4--> Mn2+Before balancing charges we have khổng lồ balance atoms. What to vày with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can địa chỉ cửa hàng H+on the left and H2O on the right:MnO4-+ H+-> Mn2++ H2OUsing simple balancing by inspection we will địa chỉ cửa hàng two coefficients khổng lồ balance atoms:MnO4-+8H+-> Mn2++4H2OOnce the atoms are balanced it is time lớn balance charge with electrons - there is +7 charge on the left side of the equation & +2 on the right side. To balance charges we have to địa chỉ cửa hàng 5 electrons on the left:MnO4-+8H++5e--> Mn2++4H2OAt the moment we have two balanced half reactions. Here comes the final trick - we địa chỉ these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to lớn use just numbers of electrons, although if often means you will have to find lowest coefficients later). To lớn have five electrons in both equations we have khổng lồ multiply first equation by 5:5Fe2+->5Fe3++5e-and when we địa chỉ cửa hàng both half reactions we will getMnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e-Canceling out the electrons:MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+And the reaction is balanced.Balancing hydrogen & oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can địa chỉ H+on the side where there is oxygen excess và water on the second, just as we did in the above example. But we can also use OH-and water to vì the trick, for example half reaction:ClO--> Cl-is not balanced, but once we add water & OH-:ClO-+ H2O -> Cl-+2OH-we have not only balanced atoms but we are also ready lớn balance charge by adding two electrons on the left:ClO-+ H2O +2e--> Cl-+2OH-and the half reaction is ready khổng lồ be used. General rule says that if the reaction takes place in acidic conditions we use water và H+to balance oxygens, & if the reaction takes place in basic conditions - we use OH-and water. Don't worry if it looks lượt thích the reaction produces H+in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.
Rishi Sharma
ccevents.vn Faculty 646 Points
2 years ago
Xem thêm: Thế Vận Hội Olympic Hiện Đại Được Tổ Chức Mấy Năm 1 Lần ? Sơ Lược Về Đại Hội Thể Thao Thế Giới (Thế Vận Hội
Dear Student,Please find below the solution khổng lồ your problem.2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2O FeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2O First of all, let's concentrate on what is really important - on the net ionic reaction: Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2O All other ions (K+and SO42-) are only spectators and we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method. At first glance you can see that in the reaction iron gets oxidized & permanganate gets reduced: Fe2+-> Fe3+ MnO4--> Mn2+ We will start equation balancing with balancing these half reactions - using electrons khổng lồ balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. Lớn balance charge we will địa chỉ one electron on the right side: Fe2+-> Fe3++ e- We can use electrons safely, as the final step of balancing will be electron cancellation. Now we have lớn balance permanganate reduction half reaction: MnO4--> Mn2+ Before balancing charges we have lớn balance atoms. What to vì with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can địa chỉ cửa hàng H+on the left và H2O on the right: MnO4-+ H+-> Mn2++ H2O Using simple balancing by inspection we will địa chỉ two coefficients to lớn balance atoms: MnO4-+8H+-> Mn2++4H2O Once the atoms are balanced it is time lớn balance charge with electrons - there is +7 charge on the left side of the equation & +2 on the right side. To balance charges we have to showroom 5 electrons on the left: MnO4-+8H++5e--> Mn2++4H2O At the moment we have two balanced half reactions. Here comes the final trick - we địa chỉ cửa hàng these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is lớn use just numbers of electrons, although if often means you will have lớn find lowest coefficients later). Lớn have five electrons in both equations we have lớn multiply first equation by 5: 5Fe2+->5Fe3++5e- & when we địa chỉ cửa hàng both half reactions we will get MnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e- Canceling out the electrons: MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+ and the reaction is balanced. Balancing hydrogen & oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. Lớn balance oxygen we can showroom H+on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH-and water to vì the trick, for example half reaction: ClO--> Cl- is not balanced, but once we showroom water and OH-: ClO-+ H2O -> Cl-+2OH- we have not only balanced atoms but we are also ready to balance charge by adding two electrons on the left: ClO-+ H2O +2e--> Cl-+2OH- và the half reaction is ready khổng lồ be used. General rule says that if the reaction takes place in acidic conditions we use water và H+to balance oxygens, và if the reaction takes place in basic conditions - we use OH-and water. Don't worry if it looks like the reaction produces H+in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.Thanks & Regards
Rishi Sharma
ccevents.vn Faculty 646 Points
2 years ago
Xem thêm: Combo Chăm Sóc Da Mặt Cho Nam Giới Tốt Nhất Hiện Nay, Mỹ PhẩM Cho Nam
Dear Student,Please find below the solution to your problem.2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2O FeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2O First of all, let's concentrate on what is really important - on the net ionic reaction: Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2O All other ions (K+and SO42-) are only spectators và we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method. At first glance you can see that in the reaction iron gets oxidized & permanganate gets reduced: Fe2+-> Fe3+ MnO4--> Mn2+ We will start equation balancing with balancing these half reactions - using electrons lớn balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. To lớn balance charge we will showroom one electron on the right side: Fe2+-> Fe3++ e- We can use electrons safely, as the final step of balancing will be electron cancellation. Now we have lớn balance permanganate reduction half reaction: MnO4--> Mn2+ Before balancing charges we have to lớn balance atoms. What to vày with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can địa chỉ H+on the left và H2O on the right: MnO4-+ H+-> Mn2++ H2O Using simple balancing by inspection we will địa chỉ cửa hàng two coefficients to balance atoms: MnO4-+8H+-> Mn2++4H2O Once the atoms are balanced it is time to balance charge with electrons - there is +7 charge on the left side of the equation & +2 on the right side. Khổng lồ balance charges we have to add 5 electrons on the left: MnO4-+8H++5e--> Mn2++4H2O At the moment we have two balanced half reactions. Here comes the final trick - we add these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to lớn use just numbers of electrons, although if often means you will have khổng lồ find lowest coefficients later). Lớn have five electrons in both equations we have lớn multiply first equation by 5: 5Fe2+->5Fe3++5e- and when we địa chỉ cửa hàng both half reactions we will get MnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e- Canceling out the electrons: MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+ & the reaction is balanced. Balancing hydrogen and oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can showroom H+on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH-and water to vì chưng the trick, for example half reaction: ClO--> Cl- is not balanced, but once we địa chỉ water and OH-: ClO-+ H2O -> Cl-+2OH- we have not only balanced atoms but we are also ready khổng lồ balance charge by adding two electrons on the left: ClO-+ H2O +2e--> Cl-+2OH- và the half reaction is ready khổng lồ be used. General rule says that if the reaction takes place in acidic conditions we use water and H+to balance oxygens, & if the reaction takes place in basic conditions - we use OH-and water. Don't worry if it looks lượt thích the reaction produces H+in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.Thanks and Regards
