\(\mathrm{NaClO_3}\) is used, even in spacecrafts, to produce \(\mathrm{O_2}\). The daily consumption of pure \(\mathrm{O_2}\) by a person is 492 L at 1 atm, 300 K. How much amount of \(\mathrm{NaClO_3}\), in grams, is required to produce \(\mathrm{O_2}\) for the daily consumption of a person at 1 atm, 300 K?
\(\mathrm{NaClO_3(s) + Fe(s) \to O_2(g) + NaCl(s) + FeO(s)}\)
\(R = 0.082\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}}\)
\(\mathrm{NaClO_3(s) + Fe(s) \to O_2(g) + NaCl(s) + FeO(s)}\)
\(R = 0.082\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}}\)
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Correct Answer: 2130
Solution and Explanation
Taking the reaction as printed (NaClO3 : O2 = 1 : 1), moles of NaClO3 = 20 mol.
Molar mass of NaClO3 = 23 + 35.5 + 48 = 106.5 g/mol.
Mass = 20 x 106.5 = 2130 g.
Answer: 2130
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