Question

sodium peroxide (na₂o₂) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. it works by reacting with co₂ in the air to produce sodium carbonate (na₂co₃) and o₂ according to the following reaction:
2 na₂o₂(s) + 2 co₂(g) —→ 2 na₂co₃(s) + o₂(g)
what volume, in l, of o₂ can be consumed at stp by 690 g na₂o₂? report your answer to the nearest whole number.
note: r = 8.314 \\(\frac{kpa\cdot l}{mol\cdot k}\\), stp = 1 bar pressure, 273.15 k

Explanation:

Step1: Calculate moles of \( \text{Na}_2\text{O}_2 \)

Molar mass of \( \text{Na}_2\text{O}_2 \) is \( 2\times23 + 2\times16 = 78 \, \text{g/mol} \).
Moles of \( \text{Na}_2\text{O}_2 = \frac{690 \, \text{g}}{78 \, \text{g/mol}} = 8.846 \, \text{mol} \).

Step2: Relate moles of \( \text{Na}_2\text{O}_2 \) to \( \text{O}_2 \)

From the reaction: \( 2 \, \text{mol} \, \text{Na}_2\text{O}_2 \) produces \( 1 \, \text{mol} \, \text{O}_2 \).
Moles of \( \text{O}_2 = \frac{8.846 \, \text{mol}}{2} = 4.423 \, \text{mol} \).

Step3: Use ideal gas law at STP

At STP (\( P = 1 \, \text{bar} \), \( T = 273.15 \, \text{K} \)), ideal gas law \( PV = nRT \).
\( V = \frac{nRT}{P} = \frac{4.423 \, \text{mol} \times 8.314 \, \frac{\text{kPa·L}}{\text{mol·K}} \times 273.15 \, \text{K}}{100 \, \text{kPa}} \) (since \( 1 \, \text{bar} = 100 \, \text{kPa} \)).
Calculate: \( V \approx \frac{4.423 \times 8.314 \times 273.15}{100} \approx 99 \, \text{L} \).

Answer:

99