Question

What is the mass of `"22.4 L N"_2"` at `"NTP"`?

Answer 1

`26.081g`

Explanation:

At NTP, pressure is `1atm`, temperature `293.15K`, moles `1` and volume `22.4L`.
Since the gas is at NTP, there is one mole of nitrogen in it.

Use the ideal gas law: `PV=nRT`

`n=(PV)/(RT)`

Here, `R=0.082057338LatmK^-1mol^-1`

Calculate:

`n=(1*22.4)/(0.082057338*293.15)`

`n=22.4/24.055`

`n=0.931mol`

We have `0.931mol` of nitrogen. Using the formula `"mass"="moles"*"molar mass"`:

`0.931*28.014=26.081g`

We have `26.081g` of nitrogen.

Answer 2

The mass (weight) of `"22.4 L N"_2` is `color(blue)("26.1 g N"_2`.

Explanation:

`"NTP"` is `"20"^@"C"` or `"293.15 K"` (used for gas laws), and `"1 atm"`.

Use the equation for the ideal gas law, and solve for moles `(n)`. Once you have moles, you can calculate the mass of nitrogen gas `("N"_2)` by multiplying the moles by its molar mass.

`PV=nRT`,

where `P` is pressure, `V` is volume, `n` is moles, `R` is the gas constant (varies with units used), and `T` is temperature in Kelvins.

Organize data:

Known

`P="1 atm"`

`V="22.4 L"`

`R="0.082057338 L atm K"^(−1) "mol"^(−1)"`
https://en.wikipedia.org/wiki/Gas_constant

`T="293.15 K"`

`"Molar mass of nitrogen gas (N"_2)"` `=` `"28.014 g/mol"`

Unknown

`n`

Determining moles `"N"_2`

Rearrange the ideal gas law equation to isolate `n`. Plug in the known values and solve.

`n=(PV)/(RT)`

`"nN"_2=((1color(red)cancel(color(black)("atm"))xx22.4color(red)cancel(color(black)("L"))))/((0.082057338 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(−1) "mol"^(−1)""xx293.15color(red)cancel(color(black)("K"))))="0.931 mol N"_2"`

Mass (weight) of `"N"_2`

`0.931color(red)cancel(color(black)("mol N"_2))xx(28.014"g N"_2)/(1color(red)cancel(color(black)("mol N"_2)))="26.1 g N"_2"`