Volume of O 2 obtained at 2 a t m & 546 K , by the complete decomposition of 8.5 g N a N O 3 is ( 2 NaNO 3 → 2 NaNO 2 + O 2 )

Question

Volume of O 2 ​ obtained at 2 a t m & 546 K , by the complete decomposition of 8.5 g N a N O 3 ​ is ( 2 NaNO 3 ​ → 2 NaNO 2 ​ + O 2 ​ )

TestHub · Accepted Answer

To determine the volume of O₂ obtained, we follow these steps:

1. Calculate moles of NaNO₃:

Molar mass of NaNO₃ = $23 + 14 + 3 \times 16 = 85 g / m o l$

Moles of NaNO₃ = $\frac{8.5 g}{85 g / m o l} = 0.1 m o l$

2. Determine moles of O₂ from the balanced reaction:

$2 NaNO_{3} \to 2 NaNO_{2} + O_{2}$

From the stoichiometry, 2 moles of NaNO₃ produce 1 mole of O₂.

Moles of O₂ = $\frac{0.1 m o l N a N O _{3}}{2} = 0.05 m o l O_{2}$

3. Apply the Ideal Gas Law to find the volume of O₂:

The Ideal Gas Law is $P V = n RT$ , so $V = \frac{n RT}{P}$ .

Given: $n = 0.05 m o l$ , $R = 0.0821 L \cdot a t m / (m o l \cdot K)$ , $T = 546 K$ , $P = 2 a t m$ .

$V = \frac{( 0.05 m o l ) ( 0.0821 L \cdot a t m / ( m o l \cdot K )) ( 546 K )}{2 a t m}$

$V = \frac{2.2413 L \cdot a t m}{2 a t m} = 1.12065 L$

Rounding to two decimal places, $V = 1.12 L$ .

The final answer is $1.12 L$ .

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