To balance the reaction`Na+O_2rarrNa_2O` the number of sodium and disodium oxide are

Options

A. 2 and 2 respectively
B. 4 and 2 respectively
C. 2 and 4 respectively
D. 3 and 5 respectively

Answer

To balance the reaction \( \text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O} \), we need to ensure that the number of each type of atom is the same on both sides of the equation. Here’s a step-by-step breakdown of how to do this:

### Step 1: Write the unbalanced equation
The unbalanced equation is:
\[ \text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O} \]

### Step 2: Count the number of atoms on each side
- On the left side (reactants):
- Sodium (Na): 1
- Oxygen (O): 2

- On the right side (products):
- Sodium (Na): 2 (from \( \text{Na}_2\text{O} \))
- Oxygen (O): 1 (from \( \text{Na}_2\text{O} \))

### Step 3: Balance sodium (Na)
Since there are 2 sodium atoms in the product (\( \text{Na}_2\text{O} \)), we need to have 2 sodium atoms on the reactant side. We can do this by placing a coefficient of 2 in front of Na:
\[ 2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O} \]

### Step 4: Count the atoms again
Now, let's count the atoms again:
- Left side:
- Sodium (Na): 2
- Oxygen (O): 2

- Right side:
- Sodium (Na): 2
- Oxygen (O): 1

### Step 5: Balance oxygen (O)
Now we see that we have 2 oxygen atoms on the left side and only 1 on the right side. To balance the oxygen, we need to adjust the coefficient in front of \( \text{Na}_2\text{O} \). Since \( \text{O}_2 \) has 2 oxygen atoms, we can place a coefficient of 2 in front of \( \text{Na}_2\text{O} \):
\[ 2\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \]

### Step 6: Final check
Now, let's check the final counts:
- Left side:
- Sodium (Na): 4 (from \( 2 \times 2 \))
- Oxygen (O): 2

- Right side:
- Sodium (Na): 4 (from \( 2 \times 2 \))
- Oxygen (O): 2 (from \( 2 \times 1 \))

Both sides are now balanced.

### Conclusion
The balanced equation is:
\[ 4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O} \]
Thus, the number of sodium (Na) is 4 and the number of di-sodium oxide (\( \text{Na}_2\text{O} \)) is 2.