NaClO$$_3$$ is used, even in spacecrafts, to produce O$$_2$$. The daily consumption of pure O$$_2$$ by a person in 492 L at 1 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) $$\rightarrow$$ O$$_2$$(g) + NaCl(s) + FeO(s)
R = 0.082 L atm mol$$^{-1}$$ K$$^{-1}$$

First we note that the volume of oxygen required by the person is given as 492 L at a pressure of 1 atm and a temperature of 300 K. To convert this volume into the corresponding amount (in moles) of O2, we use the Ideal-Gas equation, which is stated as $$PV = nRT$$ where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is absolute temperature.

Re-arranging the equation to obtain $$n$$ gives $$n = \dfrac{PV}{RT}\;.$$

Substituting the numerical values, we have

$$n = \dfrac{(1\;\text{atm})(492\;\text{L})}{(0.082\;\text{L atm mol}^{-1}\text{K}^{-1})(300\;\text{K})}\;.$$

Simplifying the denominator first, $$0.082\times 300 = 24.6\;.$$

Hence,

$$n = \dfrac{492}{24.6} = 20\;\text{mol}\;.$$

Thus the daily requirement of pure oxygen for the person is $$20$$ moles.

Now we turn to the chemical reaction that produces the oxygen:

$$\text{NaClO}_3(s) + \text{Fe}(s) \;\rightarrow\; \text{O}_2(g) + \text{NaCl}(s) + \text{FeO}(s)\;.$$

We check the atom balance in this equation. On the left there are 1 Na, 1 Cl, 1 Fe and 3 O atoms. On the right there are 1 Na (in NaCl), 1 Cl (in NaCl), 1 Fe (in FeO) and $$2+1=3$$ oxygen atoms (two in O2 and one in FeO). Because every element balances exactly, the coefficients in the written equation are already the smallest whole numbers. Hence one mole of NaClO3 produces one mole of O2.

Therefore, to obtain $$20$$ moles of O2, we need exactly $$20$$ moles of NaClO3.

Next we convert these moles into grams. The molar mass of NaClO3 is calculated as

$$M(\text{NaClO}_3) = M(\text{Na}) + M(\text{Cl}) + 3M(\text{O}) = 23 + 35.5 + 3(16) = 23 + 35.5 + 48 = 106.5\;\text{g mol}^{-1}\;.$$

Multiplying the molar mass by the number of moles gives the required mass:

$$m = n\,M = 20 \times 106.5\;\text{g} = 2130\;\text{g}\;.$$

So, the answer is $$2130\;\text{g}\;.$$