Question

${\mathbf{NaClO}}_{\mathbf{3}}$ is used, even in spacecrafts, to produce ${\mathbf{O}}_{\mathbf{2}}$. The daily consumption of pure ${\mathbf{O}}_{\mathbf{2}}$ by a person is 492 L at 1 atm, 300 K. How much amount of ${\mathbf{NaClO}}_{\mathbf{3}}$, in grams, is required to produce ${\mathbf{O}}_{\mathbf{2}}$ for the daily consumption of a person at 1 atm, 300 K?

$$\begin{gathered} {\mathrm{NaClO}}_3\left(\mathrm{s}\right)+\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+\mathrm{FeO}\left(\mathrm{s}\right)+\mathrm{NaCl}\left(\mathrm{s}\right) \\ \mathrm{R}=0.082\mathrm{L}\mathrm{atm}{\mathrm{mol}}^{–1}{\mathrm{K}}^{–1} \end{gathered}$$

Answer 1

Step 1: Given data

Given equation- ${\mathrm{NaClO}}_3\left(\mathrm{s}\right)+\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+\mathrm{FeO}\left(\mathrm{s}\right)+\mathrm{NaCl}\left(\mathrm{s}\right)$

$\mathrm{R}=0.082\mathrm{L}\mathrm{atm}{\mathrm{mol}}^{–1}{\mathrm{K}}^{–1}$, Temperature= $300\mathrm{K}$, Pressure= $1\mathrm{atm}$, Volume= $492\mathrm{L}$

Step 2: Calculating moles of ${\mathbf{O}}_{\mathbf{2}}$

${\mathrm{NaClO}}_3\left(\mathrm{s}\right)+\mathrm{Fe}\left(\mathrm{s}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+\mathrm{FeO}\left(\mathrm{s}\right)+\mathrm{NaCl}\left(\mathrm{s}\right)$

From the above equation we can see that,

$\mathrm{Moles}\mathrm{of}{\mathrm{NaClO}}_3=\mathrm{Moles}\mathrm{of}{\mathrm{O}}_2$

$\therefore$Mole of ${\mathrm{O}}_2$

$$\begin{gathered} =\frac{\mathrm{PV}}{\mathrm{RT}} \\ =\frac{1\times 492}{0.082\times 300} \\ =\frac{492}{24.6} \\ =20\mathrm{moles} \end{gathered}$$

Step 3: Calculating the amount of ${\mathbf{NaClO}}_{\mathbf{3}}$

Molar mass of 1 mol of ${\mathrm{NaClO}}_3$is

$$\begin{gathered} =23+35.5+3\times 16 \\ =106.5\mathrm{g} \end{gathered}$$

$\therefore$Mass of 20 moles of ${\mathrm{NaClO}}_3$is

$$\begin{gathered} =\left(23+35.5+3\times 16\right)\times 20\left[\because \mathrm{Moles}\mathrm{of}{\mathrm{NaClO}}_3=\mathrm{Moles}\mathrm{of}{\mathrm{O}}_2\right] \\ =106.5\times 20 \\ =2130\mathrm{g} \end{gathered}$$

Therefore, amount of ${\mathrm{NaClO}}_3$required is $2130\mathrm{g}$