# Ammonia and oxygen produce nitrogen dioxide and water. What volume of nitrogen dioxide gas will be produced if 30.5 grams of ammonia is reacted with excess oxygen?

Apr 1, 2016

#### Answer:

40.7 L of ${\text{NO}}_{2}$ will be produced at STP.

#### Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

$\text{4NH"_3 + "7O"_2 → "4NO"_2 + "6H"_2"O}$

2. Convert grams of ${\text{NH}}_{3}$ to moles of ${\text{NH}}_{3}$

The molar mass of ${\text{NH}}_{3}$ is 17.03 g/mol.

${\text{moles of NH"_3 = 30.5 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH_3"_3)))) = "1.791 mol NH}}_{3}$

Step 3. Convert moles of ${\text{NH}}_{3}$ to moles of ${\text{NO}}_{2}$**

The molar ratio of ${\text{NO}}_{2}$ to ${\text{NH}}_{3}$ is ${\text{4 mol NO"_2:"4 mol NH}}_{3}$.

1.791 color(red)(cancel(color(black)("mol NH"_3))) × ("4 mol NO"_2)/(4 color(red)(cancel(color(black)("mol NH"_3)))) = "1.791 mol NO"_2

4. Use the Ideal Gas Law to calculate the volume of ${\text{NO}}_{2}$.

The Ideal Gas Law is

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}}$, where

• $P$ = the pressure of the gas,
• $V$ = the volume of the gas,
• $n$ = the number of moles of the gas,
• $R$ = the universal gas constant
• $T$ = the temperature of the gas

You don't give the pressure or temperature of the gas, so I will assume STP (100 kPa and 0 °C).

We can rearrange the Ideal Gas Law to get

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} V = \frac{n R T}{P} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

V = (1.791 color(red)(cancel(color(black)("mol"))) ×8.314 color(red)(cancel(color(black)("kPa")))"·L·"color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(100 color(red)(cancel(color(black)("kPa")))) = "40.7 L"

The volume of ${\text{NO}}_{2}$ is 40.7 L.

Here's a useful video on mass-volume calculations.