\(\mathrm{NaClO}_{3}\) is used, even in spacecraft, to produce \(\mathrm{O}_{2}\). The daily consumption of pure \(\mathrm{O}_{2}\) by a person is 492 L at \(1 \mathrm{~atm}, 300 \mathrm{~K}\). How much amount of \(\mathrm{NaClO}_{3}\), in grams, is required to produce \(\mathrm{O}_{2}\) for the daily consumption of a person at \(1 \mathrm{~atm}, 300 \mathrm{~K}\) ? \(\mathrm{NaClO}_{3}(\mathrm{~s})+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{O}_{2}(\mathrm{~g})+\mathrm{NaCl}(\mathrm{~s})+\mathrm{FeO}(\mathrm{s})\) \(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Some Basic Concepts of Chemistry

NaClO_{3}

is used, even in spacecraft, 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

NaClO_{3}

, in grams, is required to produce

O_{2}

for the daily consumption of a person at

1 atm, 300 K

NaClO_{3} (s) + Fe (s) \to O_{2} (g) + NaCl (s) + FeO (s)

R = 0.082 L atm mol^{- 1} K^{- 1}

Answer: 2130

As written,

NaClO X_{3} + Fe O X_{2} + NaCl + FeO

is already balanced, so one mole of

NaClO X_{3}

releases one mole of

O X_{2}

. The daily oxygen is

n = \frac{P V}{R T} = \frac{1 \times 492}{0.082 \times 300} = 20

mol, which needs

20

mol of

NaClO X_{3}

. With molar mass

23 + 35.5 + 3 (16) = 106.5

g/mol, the mass required is

20 \times 106.5 = 2130

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer