# Calculate the volume of chlorine at 120 kPa and 20 °C that is required to convert 7 kg iron to ferric chloride?

May 20, 2016

The volume of chlorine is ${\text{372 m}}^{3}$.

#### Explanation:

The balanced equation is

${\text{2Fe + 3Cl"_2 → "2FeCl}}_{3}$

1. Calculate the moles of Fe

$\text{Moles of Fe" = 7000 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = "125.4 mol Fe}$

2. Calculate moles of ${\text{Cl}}_{2}$

$\text{Moles of Cl"_2 = 125.4 color(red)(cancel(color(black)("mol Fe"))) × ("3 mol Cl"_2)/(2 color(red)(cancel(color(black)("mol Fe")))) = "188.0 mol Cl"_2}$

3. Calculate the volume of ${\text{Cl}}_{2}$

We can use the Ideal Gas Law:

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

Rearrangement gives us

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

n = $\text{188.0 mol}$; $\textcolor{w h i t e}{m m m m m m m m m l l} R \text{= 8.314 kPa·L·K"^"-1""mol"^"-1}$;
$T = \text{(20 + 273.15) K = 293.15 K}$; $P = \text{120 kPa}$
V = (188.0 color(red)(cancel(color(black)("mol"))) × 8.314 color(red)(cancel(color(black)("kPa")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 293.15 color(red)(cancel(color(black)("K"))))/(120 color(red)(cancel(color(black)("kPa")))) = "3720 L" = "372 m"^3
The volume of chlorine is ${\text{372 m}}^{3}$.