$$ CuTHNO \rightarrow Cu ( NO _ { 3 } ) + NO _ { 2 } + H _ { 2 } O $$

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To balance the chemical equation $$ CuTHNO \rightarrow Cu(NO_3) + NO_2 + H_2O $$, we will follow these steps:

1. **Identify the compounds and their elements**: 
   - Reactants: $$ CuTHNO $$
   - Products: $$ Cu(NO_3) $$, $$ NO_2 $$, $$ H_2O $$

2. **List the number of atoms for each element on both sides**:
   - On the left (reactants):
     - Cu: 1
     - T: 1
     - H: 1
     - N: 1
     - O: 1
   - On the right (products):
     - Cu: 1 (from $$ Cu(NO_3) $$)
     - N: 3 (2 from $$ NO_2 $$ and 1 from $$ Cu(NO_3) $$)
     - O: 5 (3 from $$ Cu(NO_3) $$ and 2 from $$ NO_2 $$ and 1 from $$ H_2O $$)
     - H: 2 (from $$ H_2O $$)

3. **Balance the equation**:
   - Start with the element that appears in the least number of compounds. In this case, we can start with $$ H $$ and $$ N $$.
   - Adjust coefficients to balance the number of atoms for each element.

Let's balance the equation step by step:

- We have 1 $$ H $$ on the left and 2 $$ H $$ on the right. We can place a coefficient of 2 in front of $$ H_2O $$:
  $$
  CuTHNO \rightarrow Cu(NO_3) + NO_2 + 2H_2O
  $$

- Now, let's recount the atoms:
  - On the left:
    - Cu: 1
    - T: 1
    - H: 1
    - N: 1
    - O: 1
  - On the right:
    - Cu: 1
    - N: 3 (1 from $$ Cu(NO_3) $$ and 2 from $$ NO_2 $$)
    - O: 7 (3 from $$ Cu(NO_3) $$, 2 from $$ NO_2 $$, and 2 from $$ H_2O $$)
    - H: 4 (from $$ 2H_2O $$)

Next, we need to balance $$ N $$ and $$ O $$.

- To balance $$ N $$, we can adjust the coefficient of $$ NO_2 $$ to 2:
  $$
  CuTHNO \rightarrow Cu(NO_3) + 2NO_2 + 2H_2O
  $$

Now, let's recount the atoms again:
- On the left:
  - Cu: 1
  - T: 1
  - H: 1
  - N: 1
  - O: 1
- On the right:
  - Cu: 1
  - N: 5 (1 from $$ Cu(NO_3) $$ and 4 from $$ 2NO_2 $$)
  - O: 8 (3 from $$ Cu(NO_3) $$, 4 from $$ 2NO_2 $$, and 2 from $$ 2H_2O $$)
  - H: 4 (from $$ 2H_2O $$)

Finally, we can adjust the coefficient of $$ CuTHNO $$ to balance $$ N $$ and $$ O $$:
$$
2CuTHNO \rightarrow 2Cu(NO_3) + 4NO_2 + 4H_2O
$$

Now, let's recount the atoms:
- On the left:
  - Cu: 2
  - T: 2
  - H: 2
  - N: 2
  - O: 2
- On the right:
  - Cu: 2
  - N: 6 (2 from $$ 2Cu(NO_3) $$ and 4 from $$ 4NO_2 $$)
  - O: 12 (6 from $$ 2Cu(NO_3) $$, 8 from $$ 4NO_2 $$, and 4 from $$ 4H_2O $$)
  - H: 8 (from $$ 4H_2O $$)

The final balanced equation is:
$$
2CuTHNO \rightarrow 2Cu(NO_3) + 4NO_2 + 4H_2O
$$

This equation is now balanced with respect to all elements.
The balanced chemical equation is: $$ 2CuTHNO \rightarrow 2Cu(NO_3) + 4NO_2 + 4H_2O $$

\( CuTHNO \rightarrow Cu ( NO _ { 3 } ) + NO _ { 2 } + H _ { 2 } O \)

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