Question

The Russian Mir space station used a chemical oxygen generator system to make oxygen for the crew. The system ignited a tube of solid lithium perchlorate (LiClO4) to make oxygen and lithium chloride (LiCl): LiClO4 (s)  2O2 (g) + LiCl (s) If you have 300 g of LiClO4, then… (a) How many moles of oxygen gas will be generated?
(b) The station’s standard operating conditions are a pressure of 101.5 kPa and a temperature of 21°C. Based on this information and your answer to part (a), how many liters of oxygen gas will be generated? (Show your work.)

Answer 1

300 g of LiClO₄ produce 5.64 moles of O₂, which is equivalent to 136 L of O₂ at 101.5 kPa and 21 °C.

Let's consider the following balanced equation.

LiClO₄(s) ⇒ 2 O₂(g) + LiCl(s)

We can calculate the moles of O₂ produced from 300 g of LiClO₄ using the following relationships.

• The molar mass of LiClO₄ is 106.39 g/mol.

• The molar ratio of LiClO₄ to O₂ is 1:2.

$300 g LiClO_4 \times \frac{1molLiClO_4}{106.39gLiClO_4} \times \frac{2molO_2}{1molLiClO_4} = 5.64molO_2$

Then, we will convert 101.5 kPa to atm using the conversion factor 1 atm = 101.325 kPa.

$101.5 kPa \times \frac{1 atm}{101.325kPa} = 1.002atm$

Next, we will convert 21 °C to Kelvin using the following expression.

$K = \° C + 273.15 = 21\° C + 273.15 = 294 K$

Finally, we will calculate the volume occupied by 5.64 moles of oxygen using the ideal gas equation.

$P \times V = n \times R \times T\\\\V = \frac{n \times R \times T}{P} = \frac{5.64 mol \times (\frac{0.082atm.L}{mol.K} ) \times 294K}{1.002atm} = 136 L$

300 g of LiClO₄ produce 5.64 moles of O₂, which is equivalent to 136 L of O₂ at 101.5 kPa and 21 °C.

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