# Question ca67b

Jun 1, 2016

You need 7.20 g of ammonia.

#### Explanation:

We need to use the stoichiometry of the reaction. The balanced equation is:

$\text{4NH"_3 + "5O"_2 → "4NO" + "6H"_2"O}$

Step 1. Calculate the moles of ${\text{O}}_{2}$.

We can use the Ideal Gas Law:

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

We can rearrange this to give

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

STP = 1 bar and 0 °C.

 n = (1 color(red)(cancel(color(black)("bar"))) × 12.0 color(red)(cancel(color(black)("L"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K")))) = "0.5284 mol"#

Step 2. Calculate the moles of ${\text{NH}}_{3}$.

The balanced equation tells us that 4 mol of ${\text{NH}}_{3}$ are required for every 5 mol of ${\text{O}}_{2}$. So,

${\text{Moles of NH"_3 = 0.5284 cancel("mol O"_2) × ("4 mol NH"_3)/(5 cancel("mol O"_2)) = "0.4227 mol NH}}_{3}$

Step 3. Calculate the mass of ${\text{NH}}_{3}$

${\text{Mass of NH"_3 = 0.4227 color(red)(cancel(color(black)("mol NH"_3))) × ("17.03 g NH"_3)/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "7.20 g NH}}_{3}$