91Ó°ÊÓ

That are given sentences are :

Calculate the mass of sodium hypochlorite$\left(\mathrm{NaClO}\right)$

Given volume of bleach solution is$1.42\mathrm{gal}$

Convert gal to$\mathrm{mL}$is as follows:

$$\begin{gathered} =1.42\mathrm{gal}×\frac{3785.41\mathrm{mL}}{1\mathrm{gal}} \\ =5375.3\mathrm{mL} \end{gathered}$$

Density of bleach solution $=1.08\mathrm{g}/\mathrm{mL}$

Calculate the mass of sodium hypochloritelocalid="1652772295803" $\left(\mathrm{NaClO}\right)$in the bleach solution is localid="1652772304900" $282\mathrm{g}$.

That are given sentences are : (Part a)

Sodium hypochlorite

(a)

Sodium hypochiorite has an ionic bond between ${\mathrm{Na}}^{+}$and $\mathrm{ClO}°$ions. For this reason, this is an ionic compound

That are given equations are : (b)

Calculate the mass by volume percent by using the following expression.

Calculate the mass by volume percent by using the following expression.

$$\begin{gathered} \text{Mass percent}(\mathrm{m}/\mathrm{v})=\frac{\text{Solute mass}}{\text{Solution volume}}×100\% \\ =\frac{282\mathrm{g}}{5375.3\mathrm{mL}}×100\% \\ =5.25\% \end{gathered}$$

Therefore, the mass/volume percent (m/v) of sodium hypochlorite in the bleach solution is$5.25\%$$(\mathrm{m}/\mathrm{v})$

That are given equations are : (c)

A bleach solution is prepared by bubbling chlorine gas

(c)

A bleach solution is prepared by bubbling chlorine gas into a solution of sodium hydroxide to produce liquid water and an aqueous solution of sodium hypochlorite and sodium chloride. The balanced chemical equation of the reaction is as follows:

${\mathrm{Cl}}_2(\mathrm{g})+2\mathrm{NaOH}\left(\mathrm{aq}\right)âŸ\P \mathrm{NaCl}\left(\mathrm{aq}\right)+\mathrm{NaClO}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)$

That are given equations are : (d)

Calculate the moles of sodium hypochlorite in the bleach solution are as follows:

The balanced reaction is,

${\mathrm{Cl}}_2(\mathrm{g})+2\mathrm{NaOH}\left(\mathrm{aq}\right)âŸ\P \mathrm{NaCl}\left(\mathrm{aq}\right)+\mathrm{NaClO}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Mass of sodium hypochiorite in the bleach solution is$282\mathrm{g}$.

Calculate the moles of sodium hypochlorite in the bleach solution are as follows:

$$\begin{gathered} \text{Moles}=\frac{\text{mass of}\mathrm{NaClO}}{\text{Molar mass of}\mathrm{NaClO}} \\ =\frac{282\mathrm{g}}{74.44\mathrm{g}/\mathrm{mol}} \\ =3.79 \end{gathered}$$

According to the balance chemical equation,

Mole ratio of ${\mathrm{Cl}}_2$and $\mathrm{NaClO}$are in$1:1$ ratio.

That are given sentences are :

Therefore, volume occupied by $3.79$moles of Cl2 gas at STP is,

So, moles of chlorine gas required are$3.79$
moles.

At STP, volume occupied by 1 mole of $\mathrm{Cl}2$gas is $22.4\mathrm{L}$

Therefore, volume occupied by$3.79$moles of Cl2 gas at STP is,

$$\begin{gathered} =3.79\mathrm{mol}×\frac{22.4\mathrm{L}}{1\mathrm{mol}} \\ =84.9\mathrm{L} \end{gathered}$$

Therefore,$84.9\mathrm{L}$of chlorine gas at STP are required to produce one bottle of bleach.