Balance the following redox reaction by inserting the appropriate coefficients. so4^2- + nh3 = so3^2- + h2o + n2

Equivalent reaction equation :

3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂ + 3H₂O

In equalizing the redox reaction we can use the oxidation number method or half the reaction of the ions

Oxidation number method  :

1. Determine the reducing agent and oxidizing agent

2. Equate the element that experiences a redox reaction

3. Determine the amount of reduction / increase in oxidation number (number of atoms x change in oxidation number)

4. Equate the number of changes in the oxidation number by giving a coefficient

5. equal charge (H + for acidic situations and OH- for alkaline conditions)

6. Balance the H atom with the addition of H₂O

From the reaction :

SO₄²⁻ + NH₃ ---> SO₃²⁻ + H₂O + N₂

1. SO₄²⁻  ---> SO₃²⁻ = reduction (for S : +6 to +4)

NH₃ --->  N₂ = oksidation (for N : -3 to 0)

2. add coefficient  to equate atom N

SO₄²⁻ + 2NH₃ ---> SO₃²⁻ + N₂

3. SO₄²⁻ + 2NH₃ ---> SO₃²⁻ + N₂

  |_+6___2 x 3__+4_|

                  |_-66 x 1___0_|

4. giving a coefficient

3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂

5. charge already the same (left = -6, right = -6)

6. Balance the H atom with the addition of H₂O

3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂ + 3H₂O > Balance

an oxidation-reduction reaction

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Keywords : Balance, redox, coefficients,Oxidation number, half reaction

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.