Volume of `O_(2)` obtained at 2 atm & 546K, by the complete decomposition of 8.5 g `NaNO_(3)` is

Options

A. 2.24 lt.
B. 1.12 lt.
C. 0.84 lt.
D. 0.56 lt.

Answer

To solve the problem of finding the volume of \( O_2 \) obtained from the complete decomposition of 8.5 g of \( NaNO_3 \) at 2 atm and 546 K, we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of \( NaNO_3 \).

The balanced equation for the decomposition of sodium nitrate (\( NaNO_3 \)) is:
\[ 2 \, NaNO_3 \rightarrow 2 \, NaNO_2 + O_2 \]
This means that 2 moles of sodium nitrate produce 1 mole of oxygen gas.

### Step 2: Calculate the molar mass of \( NaNO_3 \).

The molar mass of \( NaNO_3 \) can be calculated as follows:
- Sodium (Na): 23 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Adding these together:
\[ \text{Molar mass of } NaNO_3 = 23 + 14 + 48 = 85 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of \( NaNO_3 \) in 8.5 g.

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
We can find:
\[ \text{Number of moles of } NaNO_3 = \frac{8.5 \, \text{g}}{85 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 4: Determine the number of moles of \( O_2 \) produced.

From the balanced equation, we know that 2 moles of \( NaNO_3 \) produce 1 mole of \( O_2 \). Therefore, 0.1 moles of \( NaNO_3 \) will produce:
\[ \text{Moles of } O_2 = \frac{0.1}{2} = 0.05 \, \text{mol} \]

### Step 5: Use the ideal gas law to find the volume of \( O_2 \).

The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (2 atm)
- \( V \) = volume (unknown)
- \( n \) = number of moles of gas (0.05 mol)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (546 K)

Rearranging the formula to solve for \( V \):
\[ V = \frac{nRT}{P} \]

Substituting the values:
\[ V = \frac{0.05 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 546 \, \text{K}}{2 \, \text{atm}} \]

### Step 6: Calculate the volume.

Calculating the numerator:
\[ 0.05 \times 0.0821 \times 546 = 2.24373 \, \text{L·atm} \]

Now, divide by the pressure:
\[ V = \frac{2.24373}{2} = 1.121865 \, \text{L} \]

Rounding to two decimal places, the volume of \( O_2 \) is approximately:
\[ V \approx 1.12 \, \text{L} \]

### Final Answer:
The volume of \( O_2 \) obtained at 2 atm and 546 K from the complete decomposition of 8.5 g of \( NaNO_3 \) is approximately **1.12 liters**.

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