NaClO₃ is used, even in spacecrafts, to produce O₂. The daily consumption of pure O₂ by a person is 492 L at 1 a t m , 300 K . How much NaClO₃, in grams, is required to produce O₂ for the daily consum

Question

NaClO₃ is used, even in spacecrafts, to produce O₂. The daily consumption of pure O₂ by a person is 492 L at $1 a t m, 300 K$ . How much NaClO₃, in grams, is required to produce O₂ for the daily consumption of a person at $1 a t m, 300 K$ ?

NaClO₃(s) + Fe(s) → O₂(g) + NaCl(s) + FeO(s)

$R = 0.082 L a t mm o l^{- 1} K^{- 1}$

TestHub · Accepted Answer

Question Explanation: Finding the mass of reactant needed for a specific volume of gas.

Concept: Ideal Gas Law and Stoichiometry

Solution: First, calculate the moles of oxygen required using the ideal gas equation ( $P V = n RT$ ). Then, use the $1 : 1$ molar ratio from the reaction to find the mass of NaClO₃ required.

Moles of O₂ consumed:

$n = \frac{P V}{RT} = \frac{1 \times 492}{0.082 \times 300} = 20 m o l$

Moles of NaClO₃ required:

$n_{N a Cl O_{3}} = 20 m o l$

Mass of NaClO₃:

$M a ss = 20 \times 106.5 = 2130 g m$

Final answer: 2130

Chemistry - Ideal Gas Question with Solution | TestHub

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