`NaClO_(3)` is used even is spacecrafts to produce `O_(2)` . The daily consumption fo pure `O_(2)` by a person is 492 L at atm , 300k .How much amount of `NaClO_(3)`, in grams is required to produce `O_(2)` for the daily consumption of a person at 1 atm 300K ?_________.
`NaClO_(3)(s) +Fe(s) to O_(2)(g) +NaCl_(s) +FeO(s)`
`R = 0.082L atm mol^(-) K^(-1)`

Explanation

`NaClO_(3)(s) + Fe(s) rarr O_(2)(g) + NaCl(s) + FeO(s)`
Number of moles of `O_(2)` = Number of moles of `NaClO_(3)`
Number of moles of `O_(2)(n) = PV//RT = 1 xx 492/0.082xx300 = 20`
Molar mass of `NaClO_(3) = 23 + 35.5 + 48 = 106.5`
Mass of `NaClO_(3) "required" = 20 xx 106.5 = 2130 g`