NaClO3 is used in spacecraft to produce O2 . The daily consumption of pure O2 by a person is 492L at 1atm, 300K. How much amount of NaClO3 (in grams) is required to produce O2 for the daily consumption of a person at 1atm, 300K? (NaClO)3(s)+Fe(s) arrow O2(g)+NaCl(s)+FeO(s) Given: R=0.082Latmmol- 1K- 1

Question

NaClO3 is used in spacecraft to produce O2 . The daily consumption of pure O2 by a person is 492L at 1atm, 300K. How much amount of NaClO3 (in grams) is required to produce O2 for the daily consumption of a person at 1atm, 300K? (NaClO)3(s)+Fe(s) arrow O2(g)+NaCl(s)+FeO(s) Given: R=0.082Latmmol- 1K- 1 (A) 21.3 (B) 1115 (C) 2130 (D) 4260

Tardigrade Admin · Accepted Answer

Moles of textNaCl textO3= moles of O2 moles of O2=( textPV/ textRT)=(1 × 492/0.082 × 300)=20 moles As molecular mass of textNaCl textO3=106.5 textNaCl textO3=20× 106.5= text2130 text g

$21.3$

$1115$

$2130$

$4260$

Moles of $NaCl O_{3} =$ moles of $O_{2}$
moles of $O_{2} = \frac{PV}{RT} = \frac{1 \times 492}{0.082 \times 300} = 20$ moles
As molecular mass of $NaCl O_{3} = 106.5$
$NaCl O_{3} = 20 \times 106.5 = 2130 g$

21.3

1115

2130

4260

Moles of NaClO3​= moles of O2​ moles of O2​=RTPV​=0.082×3001×492​=20 moles As molecular mass of NaClO3​=106.5 NaClO3​=20×106.5=2130 g

$21.3$

$1115$

$2130$

$4260$

Moles of $NaCl O_{3} =$ moles of $O_{2}$
moles of $O_{2} = \frac{PV}{RT} = \frac{1 \times 492}{0.082 \times 300} = 20$ moles
As molecular mass of $NaCl O_{3} = 106.5$
$NaCl O_{3} = 20 \times 106.5 = 2130 g$