Question

Calculate the volume that 4.5 kg(4500 g) of ethylene gas (C2H4) will occupy at STP?

Answer 1

Well, `"STP"` specifies `1*atm`, and `273.15*K`

Explanation:

And so ………….given that `PV=nRT`

`V=(nRT)/P=((4.5xx10^3*g)/(28.05*g*mol^-1)xx0.0821*(L*atm)/(K*mol)*273.15*K)/(1*atm)`

`~=3600*L`

Answer 2

The volume of `"4500 g"` of ethylene gas is `"3600 L"`.

Explanation:

Another approach:

First calculate moles of ethylene gas.

`4500color(red)cancel(color(black)("g C"_2"H"_4))xx(1"mol C"_2"H"_4)/(28.054color(red)cancel(color(black)("g C"_2"H"_4)))="160.4 mol C"_2"H"_4"`

I am leaving a couple of guard digits to reduce rounding errors. I will round the final answer to two significant figures.

At STP of `"273.15 K"` and `"1 atm"`, the molar volume of a gas is `"22.414 L/mol".`

Volume of `"C"_2"H"_4"`

Multiply mol `"C"_2"H"_4"` by the molar volume at STP.

`160.4color(red)cancel(color(black)("mol C"_2"H"_4))xx("22.414 L")/(1color(red)cancel(color(black)("mol C"_2"H"_4)))="3600 L"` rounded to two significant figures

At the current STP of `"273.15 K"` and `10^5` `"Pa"`, the molar volume of a gas is `"22.711 L/mol"`.

Volume of `"C"_2"H"_4"` at current STP

`160.4color(red)cancel(color(black)("mol C"_2"H"_4))xx("22.711 L")/(1color(red)cancel(color(black)("mol C"_2"H"_4)))="3600 L"` rounded to two significant figures