"Write and balance the equation for the reaction between nitric acid and potassium iodide. The products are potassium nitrate, iodine, nitrogen monoxide, and water."
Please explain step by step how you come to your answer please!!
Balance this Chem Equation?
2017-01-17 12:27 am
回答 (1)
2017-01-17 1:12 am
Method 1 : Oxidation number method
(1) Write the unbalanced equation as follows. There are two “HNO₃” because one of them is reduced to NO and another is converted to KNO₃. The coefficient of KI is firstly assigned to be “2” because the product I₂ contains 2 I atoms.
_HNO₃ + _HNO₃ + 2KI → _KNO₃ + _I₂ + _NO + _H₂O
(2) To balance the electrons transferred.
Oxidation: 2KI to I₂ (Oxidation number of I: -1×2 → 0), 2 electrons are lost.
Reduction: first HNO₃ to NO (Oxidation number of N: +5 → +2), 3 electrons are gained.
The LCM of 2 and 3 = 6
The coefficients of KI and I₂ should be “6” and “3” respectively, such that 6 electrons are lost.
The coefficients of both HNO₃ and NO should be “2”, such that 6 electrons are gained.
2HNO₃ + _HNO₃ + 6KI → _KNO₃ + 3I₂ + 2NO + _H₂O
(3) The coefficient of KNO₃ should be “6”, such that there are 6 K atoms on each side.
2HNO₃ + _HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + _H₂O
(4) The coefficient of second HNO₃ should be “6”, such that there are 6 N atoms on each side.
2HNO₃ + 6HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + _H₂O
Thus, 8HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + _H₂O
(5) The coefficient of H₂O should be 4, such that there are 8 H atoms on each side.
The answer: 8HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + 4H₂O
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Method 2 : Half-equations method
(1) Write the half equations for the oxidation and reduction :
Oxidation: 2I⁻ → I₂ + 2e⁻ …… [*]
Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O …… [#]
(2) [*]×3 + [#]×2, and cancel 6e⁻ on the both sides.
2NO₃⁻ + 8H⁺ + 6I⁻ → 3I₂ + 2NO + 4H₂O
(3) Add the dummy ions (i.e. 6NO₃⁻ and 6K⁺) to the both sides.
(2NO₃⁻ + 6NO₃⁻ + 8H⁺) + (6K⁺ + 6I⁻) → (6K⁺ + 6NO₃⁻) + 3I₂ + 2NO + 4H₂O
(4) Group the corresponding opposite ions into the formulae of compound.
The answer: 8HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + 4H₂O
(1) Write the unbalanced equation as follows. There are two “HNO₃” because one of them is reduced to NO and another is converted to KNO₃. The coefficient of KI is firstly assigned to be “2” because the product I₂ contains 2 I atoms.
_HNO₃ + _HNO₃ + 2KI → _KNO₃ + _I₂ + _NO + _H₂O
(2) To balance the electrons transferred.
Oxidation: 2KI to I₂ (Oxidation number of I: -1×2 → 0), 2 electrons are lost.
Reduction: first HNO₃ to NO (Oxidation number of N: +5 → +2), 3 electrons are gained.
The LCM of 2 and 3 = 6
The coefficients of KI and I₂ should be “6” and “3” respectively, such that 6 electrons are lost.
The coefficients of both HNO₃ and NO should be “2”, such that 6 electrons are gained.
2HNO₃ + _HNO₃ + 6KI → _KNO₃ + 3I₂ + 2NO + _H₂O
(3) The coefficient of KNO₃ should be “6”, such that there are 6 K atoms on each side.
2HNO₃ + _HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + _H₂O
(4) The coefficient of second HNO₃ should be “6”, such that there are 6 N atoms on each side.
2HNO₃ + 6HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + _H₂O
Thus, 8HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + _H₂O
(5) The coefficient of H₂O should be 4, such that there are 8 H atoms on each side.
The answer: 8HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + 4H₂O
====
Method 2 : Half-equations method
(1) Write the half equations for the oxidation and reduction :
Oxidation: 2I⁻ → I₂ + 2e⁻ …… [*]
Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O …… [#]
(2) [*]×3 + [#]×2, and cancel 6e⁻ on the both sides.
2NO₃⁻ + 8H⁺ + 6I⁻ → 3I₂ + 2NO + 4H₂O
(3) Add the dummy ions (i.e. 6NO₃⁻ and 6K⁺) to the both sides.
(2NO₃⁻ + 6NO₃⁻ + 8H⁺) + (6K⁺ + 6I⁻) → (6K⁺ + 6NO₃⁻) + 3I₂ + 2NO + 4H₂O
(4) Group the corresponding opposite ions into the formulae of compound.
The answer: 8HNO₃ + 6KI → 6KNO₃ + 3I₂ + 2NO + 4H₂O
收錄日期: 2021-05-01 14:16:02
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