NaClO3 is used, even in spacecrafts, to produce O2. The daily consumption of pure O2 by a person is 492 Lat 1 atm, 300 K. How much amount of NaClO3, in grams, is required to produce O2 for the daily consumption of a person at 1 atm, 300 K ? . NaClO3(s) + Fe(s) - O2(g) + NaCl(s) + FeO(s) R=0.082 L atm mol-1K-1 Given 492.365

Question

NaClO3 is used, even in spacecrafts, to produce O2. The daily consumption of pure O2 by a person is 492 Lat 1 atm, 300 K. How much amount of NaClO3, in grams, is required to produce O2 for the daily consumption of a person at 1 atm, 300 K ? . NaClO3(s) + Fe(s) -> O2(g) + NaCl(s) + FeO(s) R=0.082 L atm mol-1K-1 Given 492.365 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mole of O2 consumed (1×492/0.082×300)=20 Mole of NaClO3 required = 20 Mass of NaClO3 = 20 Ã 106.5 = 2130 gm

Mole of O2​ consumed 0.082×3001×492​=20 Mole of NaClO3​ required =20 Mass of NaClO3​=20×106.5=2130gm

Mole of $O_{2}$ consumed $\frac{1 \times 492}{0.082 \times 300} = 20$
Mole of $N a Cl O_{3}$ required $= 20$
Mass of $N a Cl O_{3} = 20 \times 106.5 = 2130 g m$