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Q. $NaClO_3$ is used, even in spacecrafts, to produce $O_2$. The daily consumption of pure $O_2$ by a person is 492 Lat 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, $300\, K$ ? ___________.
$\ce{NaClO3(s) + Fe(s) -> O2(g) + NaCl(s) + FeO(s)}$
$R=0.082\,L\,atm\,mol^{-1}K^{-1}$
Given 492.365

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Solution:

Mole of $O_2$ consumed $\frac{1\times492}{0.082\times300}=20$
Mole of $NaClO_{3}$ required $= 20$
Mass of $NaClO_{3} = 20 × 106.5 = 2130\, gm$