Question

For the combustion reaction `4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(g)`, how many liters of `"O"_2(g)` would react with `"500 L"` of `"NH"_3` at STP?

Answer 1

`"625 L O"_2`

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This is a limiting reactant problem using gases (assumed ideal).

Given `"500 L"` of `"NH"_3`, since we are at STP (presumably, `"1 atm"` and `0^@ "C"`), we can determine the molar volume of `"NH"_3` by assuming it is an ideal gas.

`PV = nRT`

`V/n = (RT)/P`

`= (("0.082057 L"cdotcancel"atm""/mol"cdotcancel"K")(273.15 cancel"K"))/(cancel"1 atm")`

`=` `"22.41 L/mol"`

Therefore, we have

`500 cancel("L NH"_3) xx "1 mol"/(22.41 cancel("L NH"_3(g)))`

`=` `"22.31 mols NH"_3`

From the reaction given, there needs to be `"5 mols O"_2(g)` for every `"4 mols NH"_3(g)`. Therefore:

`22.31 cancel("mols NH"_3) xx ("5 mols O"_2)/(4 cancel("mols NH"_3))`

`=` `"27.88 mols O"_2`

Since we are also assuming `"O"_2(g)` is an ideal gas, we utilize the same molar volume to get:

`27.88 cancel("mols O"_2(g)) xx "22.41 L"/cancel("mol O"_2(g))`

`=` `color(blue)("625 L O"_2(g))`

You could also have gone straight from the volume and used the mol ratio by ASSUMING that both `"O"_2(g)` and `"NH"_3(g)` are ideal gases:

`"500 L"` `cancel("NH"_3) xx (5 cancel"mols" "O"_2)/(4 cancel"mols" cancel("NH"_3))`

`= color(blue)("625 L O"_2(g))`