# How many mL of carbon tetrachloride are produced when 8.0 L of chlorine are allowed to react with 0.75 L of methane at STP?

Jul 2, 2016

You will get 3.2 mL of ${\text{CCl}}_{4}$ at STP.

#### Explanation:

This is a limiting reactant problem.

One way to solve it is to figure out how much product you could get from each reactant separately.

1. Write the balanced equation.

The balanced chemical equation is

$\text{CH"_4 + "4Cl"_2 → "CCl"_4 + "4HCl}$

2. Identify the limiting reactant

We can use Gay-Lussac's Law of Combining Volumes to identify the limiting reactant.

From ${\text{CH}}_{4}$:

The volume ratio is $\left({\text{1 L CCl"_4)/("1 L CH}}_{4}\right)$

${\text{Volume of CCl"_4 = 0.75 color(red)(cancel(color(black)("L CH"_4))) × ("1 L CCl"4)/(1 color(red)(cancel(color(black)("L CH"_4)))) = "0.75 L CCl}}_{4}$

From ${\text{Cl}}_{2}$:

The volume ratio is $\left({\text{1 L CCl"_4)/("4 L Cl}}_{2}\right)$.

${\text{Volume of CCl"_4 = 8.0 color(red)(cancel(color(black)("L Cl"_2))) × ("1 mol CCl"_4)/(4 color(red)(cancel(color(black)("L Cl"_2)))) = "2.0 L CCl}}_{4}$

${\text{CH}}_{4}$ gives the smaller volume of gaseous ${\text{CCl}}_{4}$, so ${\text{CH}}_{4}$ is the limiting reactant.

The theoretical yield of gaseous ${\text{CCl}}_{4}$ at STP is 0.75 L.

3. Calculate the mass of ${\text{CCl}}_{4}$

To solve this problem, we can use the Ideal Gas Law:

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

Since $n = \text{mass"/"molar mass} = \frac{m}{M}$, we can write the Ideal Gas Law as

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

We can rearrange this to get

$m = \frac{P V M}{R T}$

STP is defined at 1 bar and 0 °C.

$P = \text{1 bar}$
$V = \text{0.75 L}$
$M = \text{153.82 g/mol}$
$R = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T = \text{273.15 K}$

If ${\text{CCl}}_{4}$ were an ideal gas at STP, we would calculate that

m = (1 color(red)(cancel(color(black)("bar"))) × 0.75 color(red)(cancel(color(black)("L"))) × "153.82 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "5.08 g"

4. Calculate the volume of ${\text{CCl}}_{4}$

${\text{CCl}}_{4}$ is a liquid at STP. Its density is about 1.59 g/mL.

∴ The volume is

V = 5.08 color(red)(cancel(color(black)("g"))) × "1 mL"/(1.59 color(red)(cancel(color(black)("g")))) = "3.2 mL"