5 nano2 + 2 kmno4 + 3 h2so4 → 5 nano3 + 2 mnso4 + k2so4 + 3 h2o

     

You are both wrong. The role of H2SO4 is nothing. H⁺ remains in the same oxidation state as does SO4⁻². Mn is in the +7 state and goes to +2. (reduction) N goes from +3 to +5. You could write the equation as MnO4- + NO2- -> Mn+2 + NO3-


You are both wrong. The role of H2SO4 is nothing. H⁺ remains in the same oxidation state as does SO4⁻². Mn is in the +7 state & goes khổng lồ +2. (reduction) N goes from +3 to +5. You could write the equation as MnO4- + NO2- -> Mn+2 + NO3-

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You are both wrong. The role of H2SO4 is nothing. H⁺ remains in the same oxidation state as does SO4⁻². Mn is in the +7 state và goes khổng lồ +2. (reduction) N goes from +3 khổng lồ +5. You could write the equation as MnO4- + NO2- -> Mn+2 + NO3-


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I don’t think either of you is correct. In the reaction nitrogen (in NaNO2) is, formally, in the +3 oxidation state và in the product, it is in the +5 oxidation state so, in the process, it was oxidized. Now, consider the Mn. In KMnO4, Mn is in the +7 oxidation state. In the sản phẩm (MnSO4), it is in the +2 oxidation state & therefore Mn was reduced in the overall process. Finally, S, in H2SO4, K2SO4 và MnSO4 remains in the +6 oxidation state, unchanged. So the reductant in this reaction is NO2(1-) (and thereby is oxidized to lớn NO3(-1)) & MnO4- is the oxidant (being reduce to Mn(2+).

Bạn đang xem: 5 nano2 + 2 kmno4 + 3 h2so4 → 5 nano3 + 2 mnso4 + k2so4 + 3 h2o

Note that the reaction does not require H2SO4 & can be done under basic conditions. In this case, the reaction shown below occurs:

NaNO2 + KMnO4 + NaOH → NaNO3 + Na2MnO4 + KOH + H2O

The main difference is that in base Mn goes from a +7 khổng lồ a +6 oxidation state. Conducting the reaction is acid is just a more efficient use of KMnO4 as you get 5 electrons from it in acid, only 1 in base.


I don’t think either of you is correct. In the reaction nitrogen (in NaNO2) is, formally, in the +3 oxidation state & in the product, it is in the +5 oxidation state so, in the process, it was oxidized. Now, consider the Mn. In KMnO4, Mn is in the +7 oxidation state. In the hàng hóa (MnSO4), it is in the +2 oxidation state và therefore Mn was reduced in the overall process. Finally, S, in H2SO4, K2SO4 và MnSO4 remains in the +6 oxidation state, unchanged. So the reductant in this reaction is NO2(1-) (and thereby is oxidized to lớn NO3(-1)) và MnO4- is the oxidant (being reduce to Mn(2+).

Xem thêm: Mua Thúng Đựng Xôi Ở Đầu - Cách Ủ Xôi Nóng Lâu Để Bán Cả Buổi Ít Người Biết

Note that the reaction does not require H2SO4 and can be done under basic conditions. In this case, the reaction shown below occurs:

NaNO2 + KMnO4 + NaOH → NaNO3 + Na2MnO4 + KOH + H2O

The main difference is that in base Mn goes from a +7 khổng lồ a +6 oxidation state. Conducting the reaction is acid is just a more efficient use of KMnO4 as you get 5 electrons from it in acid, only 1 in base.

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I don’t think either of you is correct. In the reaction nitrogen (in NaNO2) is, formally, in the +3 oxidation state & in the product, it is in the +5 oxidation state so, in the process, it was oxidized. Now, consider the Mn. In KMnO4, Mn is in the +7 oxidation state. In the sản phẩm (MnSO4), it is in the +2 oxidation state & therefore Mn was reduced in the overall process. Finally, S, in H2SO4, K2SO4 và MnSO4 remains in the +6 oxidation state, unchanged. So the reductant in this reaction is NO2(1-) (and thereby is oxidized to NO3(-1)) and MnO4- is the oxidant (being reduce to lớn Mn(2+).

Xem thêm: Potassium Nitrate Fill Solution (Kno3), 150Ml, Kno3 + S + C = K2S + N2 + Co2

Note that the reaction does not require H2SO4 and can be done under basic conditions. In this case, the reaction shown below occurs:

NaNO2 + KMnO4 + NaOH → NaNO3 + Na2MnO4 + KOH + H2O

The main difference is that in base Mn goes from a +7 to a +6 oxidation state. Conducting the reaction is acid is just a more efficient use of KMnO4 as you get 5 electrons from it in acid, only 1 in base.