# Question 9a203

May 4, 2017

Here's what I got.

#### Explanation:

You know that

2overbrace("NaCl"_ ((l)))^(color(blue)("Na"_ ((l))^(+) + "Cl"_ ((l))^(-))) -> 2"Na"_ ((s)) + "Cl"_ (2(g)) $\uparrow$

In this redox reaction, sodium is being reduced to sodium metal at the cathode (the negative electrode)

${\text{Na"_ ((l))^(+) + "e"^(-) -> "Na}}_{\left(s\right)} \to$ the reduction half-reaction

Chlorine is being oxidized to chlorine gas at the anode (the positive electrode)

$2 {\text{Cl"_ ((l))^(+) -> "C"_ (2(g)) + 2"e}}^{-} \to$ the oxidation half-reaction

Now, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the redox half-reaction, which is why you have

$\left\{\begin{matrix}2 {\text{Na"_ ((l))^(+) + 2"e"^(-) -> 2"Na"_ ((s)) \\ color(white)(aaaaaa)2"Cl"_ ((l))^(+) -> "Cl"_ (2(g)) + 2"e}}^{-}\end{matrix}\right.$

So, you know that you need $1$ mole of electrons to convert $1$ mole of molten sodium cations to sodium metal.

Start by converting the mass of sodium metal to moles

1.00 * 10^3 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole Na"/(22.99 color(red)(cancel(color(black)("g")))) = 4.35 * 10^4 $\text{moles Na}$

So, you know that you must convert $4.35 \cdot {10}^{4}$ moles of molten sodium cations to sodium metal, so you can say that you're going to need $4.35 \cdot {10}^{4}$ moles of electrons.

As you know, each mole of electrons is equivalent to $9.65 \cdot {10}^{4}$ $\text{C}$, which means that the total charge needed to convert these many moles of molten sodium cations to sodium metal is equal to

4.35 * 10^4 color(red)(cancel(color(black)("moles e"^(-)))) * (9.65 * 10^4color(white)(.)"C")/(1color(red)(cancel(color(black)("mole e"^(-))))) = 4.198 * 10^9 $\text{C}$

Now, you know that

$\text{1 A" = "1 C"/"1 s}$

This means that you will need

4.198 * 10^9 color(red)(cancel(color(black)("C"))) * overbrace("1 s"/(3.00 * 10^4color(red)(cancel(color(black)("C")))))^(color(blue)(=3.00 * 10^4"A")) = 1.399 * 10^5 $\text{s}$

to produce the required amount of sodium metal. Convert this to hours to get

$1.399 \cdot {10}^{5} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{s"))) * "1 hr"/(3600color(red)(cancel(color(black)("s")))) = color(darkgreen)(ul(color(black)("38.9 hr}}}}$

Now, notice that you need $2$ moles of molten chloride anions and $2$ moles of electrons to produce $1$ mole of chlorine gas.

This means that the reaction will produce half as many moles of chlorine gas as you have moles of sodium metal, so

4.35 * 10^4 color(red)(cancel(color(black)("moles Na"))) * "1 mole Cl"_2/(2color(red)(cancel(color(black)("moles Na")))) = 2.175 * 10^4# ${\text{moles Cl}}_{2}$

To find the volume occupied by the gas at STP, use the fact that STP conditions are usually defined as a temperature of ${0}^{\circ} \text{C}$ and a pressure of $\text{1 atm}$.

Using the ideal gas law equation

$P V = n R T$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

you will have

$V = \frac{n R T}{P}$

Plug in your values to find--do not forget to convert the temperature to Kelvin!

$V = \left(2.175 \cdot {10}^{4} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (0 + 273.15)color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{V = 4.88 \cdot {10}^{5} \textcolor{w h i t e}{.} \text{L}}}}$

The answers are rounded to three sig figs, the number of sig figs you have for your values.