Question

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 daily consumption of a person at 1 atm and 300 k

Answer 1

Answer:

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Answer 2

1410 grams  of $NaClO_3$ is required to produce $O_2$ for daily consumption of a person at 1 atm and 300 K

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = 492 L

n=  moles of gas = ?

T= Temperature of the gas in kelvin = 300 K

R= Gas constant = 0.0821Latm/Kmol

$n=\frac{PV}{RT}=\frac{1\times 492}{0.0821\times 300}=20.0moles$

The balanced chemical reaction is:

$2NaClO_3\rightarrow 2NaCl+3O_2$

According to stoichiometry:

3 moles of oxygen are produced by 2 moles of $NaClO_3$

20.0 moles of oxygen are produced by=$\frac{2}{3}\times 20.0=13.3$  moles of $NaClO_3$

Mass of $NaClO_3$ =$moles\times {\text {Molar Mass}}=13.3\times 106=1410g$

 

Learn more about stoichiometry

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