# 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?

Apr 30, 2017

${\text{625 L O}}_{2}$

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

Given $\text{500 L}$ of ${\text{NH}}_{3}$, since we are at STP (presumably, $\text{1 atm}$ and ${0}^{\circ} \text{C}$), we can determine the molar volume of ${\text{NH}}_{3}$ by assuming it is an ideal gas.

$P V = n R T$

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

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

$=$ $\text{22.41 L/mol}$

Therefore, we have

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

$=$ ${\text{22.31 mols NH}}_{3}$

From the reaction given, there needs to be ${\text{5 mols O}}_{2} \left(g\right)$ for every ${\text{4 mols NH}}_{3} \left(g\right)$. Therefore:

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

$=$ ${\text{27.88 mols O}}_{2}$

Since we are also assuming ${\text{O}}_{2} \left(g\right)$ is an ideal gas, we utilize the same molar volume to get:

$27.88 \cancel{{\text{mols O"_2(g)) xx "22.41 L"/cancel("mol O}}_{2} \left(g\right)}$

$=$ $\textcolor{b l u e}{{\text{625 L O}}_{2} \left(g\right)}$

You could also have gone straight from the volume and used the mol ratio by ASSUMING that both ${\text{O}}_{2} \left(g\right)$ and ${\text{NH}}_{3} \left(g\right)$ are ideal gases:

$\text{500 L}$ cancel("NH"_3) xx (5 cancel"mols" "O"_2)/(4 cancel"mols" cancel("NH"_3))

$= \textcolor{b l u e}{{\text{625 L O}}_{2} \left(g\right)}$