Question #9a203

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You know that

`2overbrace("NaCl"_ ((l)))^(color(blue)("Na"_ ((l))^(+) + "Cl"_ ((l))^(-))) -> 2"Na"_ ((s)) + "Cl"_ (2(g))` `uarr`

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

`"Na"_ ((l))^(+) + "e"^(-) -> "Na"_ ((s)) ->` the reduction half-reaction

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

`2"Cl"_ ((l))^(+) -> "C"_ (2(g)) + 2"e"^(-) ->` 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

`{(2"Na"_ ((l))^(+) + 2"e"^(-) -> 2"Na"_ ((s))), (color(white)(aaaaaa)2"Cl"_ ((l))^(+) -> "Cl"_ (2(g)) + 2"e"^(-)) :}`

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` `"moles Na"`

So, you know that you must convert `4.35 * 10^4` moles of molten sodium cations to sodium metal, so you can say that you're going to need `4.35 * 10^4` moles of electrons.

As you know, each mole of electrons is equivalent to `9.65 * 10^4` `"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` `"C"`

Now, you know that

`"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` `"s"`

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

`1.399 * 10^5 color(red)(cancel(color(black)("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` `"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^@"C"` and a pressure of `"1 atm"`.

Using the ideal gas law equation

`PV = nRT`

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("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

you will have

`V = (nRT)/P`

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

`V = (2.175 * 10^4 color(red)(cancel(color(black)("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"))))`

`color(darkgreen)(ul(color(black)(V = 4.88 * 10^5color(white)(.)"L")))`

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