For the combustion reaction $4 {NH}_{3} (g) + 5 O_{2} (g) \to 4 NO (g) + 6 H_{2} O (g)$ $4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(g)$ , how many liters of $O_{2} (g)$ $"O"_2(g)$ would react with $500 L$ $"500 L"$ of ${NH}_{3}$ $"NH"_3$ at STP?

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For the combustion reaction $4 {NH}_{3} (g) + 5 O_{2} (g) \to 4 NO (g) + 6 H_{2} O (g)$ $4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(g)$ , how many liters of $O_{2} (g)$ $"O"_2(g)$ would react with $500 L$ $"500 L"$ of ${NH}_{3}$ $"NH"_3$ at STP?

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Answer 1 · 2017-04-30T21:51:03

${625 L O}_{2}$ $"625 L O"_2$

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

Given $500 L$ $"500 L"$ of ${NH}_{3}$ $"NH"_3$ , since we are at STP (presumably, $1 atm$ $"1 atm"$ and $0^{\circ} C$ $0^@ "C"$ ), we can determine the molar volume of ${NH}_{3}$ $"NH"_3$ by assuming it is an ideal gas.

$P V = n R T$ $PV = nRT$

$\frac{V}{n} = \frac{R T}{P}$ $V/n = (RT)/P$

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

$=$ $"22.41 L/mol"$

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:

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

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

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For the combustion reaction $4 {NH}_{3} (g) + 5 O_{2} (g) \to 4 NO (g) + 6 H_{2} O (g)$ $4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(g)$ , how many liters of $O_{2} (g)$ $"O"_2(g)$ would react with $500 L$ $"500 L"$ of ${NH}_{3}$ $"NH"_3$ at STP?

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For the combustion reaction $4 {NH}_{3} (g) + 5 O_{2} (g) \to 4 NO (g) + 6 H_{2} O (g)$ $4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(g)$ , how many liters of $O_{2} (g)$ $"O"_2(g)$ would react with $500 L$ $"500 L"$ of ${NH}_{3}$ $"NH"_3$ at STP?