In the balanced reaction : `aNO_(3)^(-)+bCI^(-)+cH^(+)rarr dNO+eCI_(2)+fH_(2)O`
`a, b, c, d, e and f` are lowest possible integers. The value of `a + b` is :

Options

A. `6`
B. `8`
C. `4`
D. `10`

Explanation

To solve the balanced reaction given in the question:

**Step 1: Write down the unbalanced reaction.**
The reaction is given as:
\[ aNO_3^{-} + bCl^{-} + cH^{+} \rightarrow dNO + eCl_2 + fH_2O \]

**Step 2: Identify the number of atoms for each element on both sides.**
- Reactants:
- Nitrogen (N): 1 from \(NO_3^{-}\)
- Chlorine (Cl): 1 from \(Cl^{-}\)
- Hydrogen (H): 1 from \(H^{+}\)
- Oxygen (O): 3 from \(NO_3^{-}\)

- Products:
- Nitrogen (N): 1 from \(NO\)
- Chlorine (Cl): 2 from \(Cl_2\)
- Hydrogen (H): 2 from \(H_2O\)
- Oxygen (O): 1 from \(NO\) and 1 from \(H_2O\)

**Step 3: Balance the nitrogen atoms.**
Since there is 1 nitrogen atom on both sides, nitrogen is already balanced.

**Step 4: Balance the oxygen atoms.**
We have 3 oxygen atoms in the reactants and 2 in the products. To balance the oxygen, we can adjust the coefficients:
- Let’s take 2 moles of \(NO\) on the product side, which gives us 2 nitrogen and 2 oxygen.
- Therefore, we need to take 2 moles of \(NO_3^{-}\) on the reactant side, which gives us 6 oxygen.

Now the equation looks like:
\[ 2NO_3^{-} + bCl^{-} + cH^{+} \rightarrow 2NO + eCl_2 + fH_2O \]

**Step 5: Update the count of oxygen and balance it.**
Now we have:
- Reactants: 6 O (from \(2NO_3^{-}\))
- Products: 2 O (from \(2NO\)) + \(f\) O (from \(H_2O\))

To balance the oxygen, we can set \(f = 4\) (since \(H_2O\) contributes 4 O):
\[ 2NO_3^{-} + bCl^{-} + cH^{+} \rightarrow 2NO + eCl_2 + 4H_2O \]

Now we have:
- Reactants: 6 O
- Products: 2 + 4 = 6 O

**Step 6: Balance the hydrogen atoms.**
We have 4 H from \(4H_2O\) on the product side. Therefore, we need 4 \(H^{+}\) on the reactant side:
\[ 2NO_3^{-} + bCl^{-} + 4H^{+} \rightarrow 2NO + eCl_2 + 4H_2O \]

**Step 7: Balance the chlorine atoms.**
We have \(b\) \(Cl^{-}\) on the reactant side and \(e\) \(Cl_2\) on the product side. Since \(Cl_2\) has 2 Cl, we can set \(e = 3\) and \(b = 6\):
\[ 2NO_3^{-} + 6Cl^{-} + 4H^{+} \rightarrow 2NO + 3Cl_2 + 4H_2O \]

**Step 8: Write the final balanced equation.**
The final balanced equation is:
\[ 2NO_3^{-} + 6Cl^{-} + 4H^{+} \rightarrow 2NO + 3Cl_2 + 4H_2O \]

**Step 9: Identify the values of a, b, c, d, e, and f.**
From the balanced equation:
- \(a = 2\)
- \(b = 6\)
- \(c = 4\)
- \(d = 2\)
- \(e = 3\)
- \(f = 4\)

**Step 10: Calculate \(a + b\).**
Now, we need to find \(a + b\):
\[ a + b = 2 + 6 = 8 \]

**Final Answer:**
The value of \(a + b\) is \(8\).

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