The acidic aqueous solution of ferrous ion forms a brown complex in the presence of NO ${_{3}}^{-}$ , by the following two steps:
[Fe(H $_{2}$ O) $_{6}$ ] $^{2 +}$ + NO ${_{3}}^{-}$ + H $^{+}$ $\to$ ..... + [Fe(H $_{2}$ O) $_{6}$ ] $^{3 +}$ + H $_{2}$ O
[Fe(H $_{2}$ O) $_{6}$ ] $^{2 +}$ + ... $\to$ .... + H $_{2}$ O
Complete and balance the equations.

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Solution

Steps of Balancing a Chemical Equation:
1. Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced.
2. What is the net charge on each side of the equation. The net charge must be the same on each side of the equation once it has been balanced.If possible, start with an element found in one compound on each side of the equation.
3. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember! To balance an equation, you change the coefficients, not the subscripts in the formulas.
4. Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last.
5. Check your work to make certain the charge on both sides of the equation is also balanced.
$3 [F e (H_{2} O)_{6}]^{2 +} + {N O}_{3}^{-} + 4 H^{+} \to N O + 3 [F e (H_{2} O)_{6}]^{3 +} + 2 H_{2} O$
$[F e (H_{2} O)_{6}]^{2 +} + N O \to [F e (H_{2} O)_{5} N O]^{2 +} + H_{2} O$

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Similar questions

{N O}_{3}^{-}

[{F e (H_{2} O)}_{6}]^{2 +} + {N O}_{3}^{-} + H^{+} ⟶ (a) + {F e (H_{2} O)}_{6}]^{3 +} + H_{2} O

[{F e (H_{2} O)}_{6}]^{2 +} + (b) ⟶ (c) + H_{2} O

[F e (H_{2} O)_{5} N O]^{2 +}

C u_{2} O + H^{+} + N O_{3}^{-} \to C u^{2 +} + N O + H_{2} O

_{2}

_{5}

^{2 +}

_{4}

{N O_{3}}^{-}

_{2}

_{4}

_{2}

_{5}

_{4}

_{2}

_{5}

_{4}

_{2}

_{5}

_{4}

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