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Standard XII

Chemistry

Introduction to Oxidation and Reduction

NaClO3 is use...

Question

$N a C l O_{3}$ is used even in spacecrafts to produce $O_{2}$ . The daily consumption of pure $O_{2}$ by a person is 492 L at $1 a t m$ and $300 K$ . How much amount of $N a C l O_{3}$ in
grams, is required to produce $O_{2}$ for the daily consumption of a person at STP?
$N a C l O_{3} (s) + F e (s) \to O_{2} (g) + N a C 1 (s) + F e O (s)$

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Solution

$N a C l O_{3} (s) + F e (s) \to O_{2} (g) + N a C 1 (s) + F e O (s)$
Moles of $N a C l O_{3}$ required = moles of $O_{2}$ produced
Moles of $O_{2}$ required
$n = \frac{P V}{R T} = \frac{1 \times 492}{0.082 \times 300} = 20 m o l$
Molar mass of $N a C 1 O_{3}$ $= 23 + 35.5 + (3 \times 16) = 106.5 g m o l^{- 1}$
$∴$ Mass of $N a C l O_{3}$ required $= 20 \times 106.5 = 2130 g$

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Similar questions

N a C l O_{3}

is used, even in spacecrafts, 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

N a C l O_{3}

, in grams, is required to produce

O_{2}

for the daily consumption of a person at 1 atm, 300 K ?

N a C l O_{3} (s) + F e (s) \to O_{2} (g) + F e O (s) + N a C l (s)

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

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Human body requires $2370 K . c a l$ of energy daily. If the heat of combustion of glucose is $- 790 K . c a l / m o l e$ the amount of glucose required for daily consumption is

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NaClO3 is used even in spacecrafts to produce O2. The daily consumption of pure O2 by a person is 492 L at 1 atm and 300 K. How much amount of NaClO3 in grams, is required to produce O2 for the daily consumption of a person at STP? NaClO3(s)+Fe(s)→O2(g)+NaC1(s)+FeO(s)

NaClO3(s)+Fe(s)→O2(g)+NaC1(s)+FeO(s) Moles of NaClO3 required = moles of O2 produced Moles of O2 required n=PVRT=1×4920.082×300=20 mol Molar mass of NaC1O3 =23+35.5+(3×16)=106.5 g mol−1 ∴ Mass of NaClO3 required =20×106.5=2130 g