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Ideal Gas Equation

N aClO 3 is u...

Question

$N a C l O_{3}$ is used, even in spacecrafts, to produce $O_{2}$ . The daily consumption of pure $O_{2}$ by a person is 492 L at 1 atm, 300 K. How much amount of $N a C l O_{3}$ , in grams, is required to produce $O_{2}$ for the daily consumption of a person at 1 atm, 300 K ?
$N a C l O_{3} (s) + F e (s) \to O_{2} (g) + F e O (s) + N a C l (s)$
$R = 0.082 {L atm mol}^{- 1} K^{- 1}$

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Solution

Mol of $N a C l O_{3}$ = mol of $O_{2}$
Mol of $O_{2} = \frac{P V}{R T} = \frac{1 \times 492}{0.082 \times 300} = 20 mol$
Molar mass of $N a C l O_{3}$ is 106.5
So, mass $= 20 \times 106.5 = 2130 g = 2.13 kg$

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