4 mol of a mixture containing one mol each of `LiNO_(3), NaNO_(3), Ca(NO_(3))_(2),` and `Mg(NO_(3))_(2)` is decomposed by strongly heating. The total number of moles of `NO_(2)(g)` evolved is (assuming complete decomposition of all the salts)

Explanation

To solve the problem, we need to analyze the decomposition reactions of each nitrate compound in the mixture and determine how many moles of \( NO_2 \) gas are produced from each.

### Step-by-Step Solution:

1. **Identify the Nitrates in the Mixture:**
The mixture contains:
- 1 mole of \( LiNO_3 \)
- 1 mole of \( NaNO_3 \)
- 1 mole of \( Ca(NO_3)_2 \)
- 1 mole of \( Mg(NO_3)_2 \)

2. **Write the Decomposition Reactions:**
- **For \( LiNO_3 \):**
\[ 2 LiNO_3 \rightarrow Li_2O + 2 NO_2 + O_2 \]
From 1 mole of \( LiNO_3 \), we get:
\[ 1 \text{ mole of } NO_2 \]

- **For \( NaNO_3 \):**
\[ 2 NaNO_3 \rightarrow 2 NaNO_2 + O_2 \]
From 1 mole of \( NaNO_3 \), we get:
\[ 0 \text{ moles of } NO_2 \]

- **For \( Ca(NO_3)_2 \):**
\[ Ca(NO_3)_2 \rightarrow CaO + 2 NO_2 + \frac{1}{2} O_2 \]
From 1 mole of \( Ca(NO_3)_2 \), we get:
\[ 2 \text{ moles of } NO_2 \]

- **For \( Mg(NO_3)_2 \):**
\[ Mg(NO_3)_2 \rightarrow MgO + 2 NO_2 + \frac{1}{2} O_2 \]
From 1 mole of \( Mg(NO_3)_2 \), we get:
\[ 2 \text{ moles of } NO_2 \]

3. **Calculate Total Moles of \( NO_2 \):**
Now, we sum the moles of \( NO_2 \) produced from each nitrate:
- From \( LiNO_3 \): \( 1 \)
- From \( NaNO_3 \): \( 0 \)
- From \( Ca(NO_3)_2 \): \( 2 \)
- From \( Mg(NO_3)_2 \): \( 2 \)

Total moles of \( NO_2 \):
\[ 1 + 0 + 2 + 2 = 5 \text{ moles of } NO_2 \]

### Final Answer:
The total number of moles of \( NO_2(g) \) evolved is **5 moles**.

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