# Question 7fac9

Jul 25, 2015

Chlorine gas will be the limiting reagent.

#### Explanation:

Start by writing the balance d chemical equation for this reaction, which is actually called the chlorination of carbon disulfide

$C {S}_{2 \left(g\right)} + \textcolor{red}{3} C {l}_{2 \left(g\right)} \to C C {l}_{4 \left(g\right)} + {S}_{2} C {l}_{2 \left(g\right)}$

Notice that you have a $1 : \textcolor{red}{3}$ mole ratio between carbond disulfide and chlorine gas. This means that, rgardless of how many moles of the former you have, you need three times more moles of the latter in order for the reaction to take place.

Automatically, chlorine gas will act as a limiting reagent, since you start with 0.235 moles instead of

0.270cancel("moles"CS_2) * (color(red)(3)" moles "Cl_2)/(1cancel("mole"CS_2)) = "0.810 moles" $C {l}_{2}$

Therefore, the formula of the limiting reagent is $C {l}_{2}$.

To get the maximum amount of carbon tetrachloride that can be produced by this reaction, use the fact that you ahve a $1 : 1$ mole ratio between carbon disulfide and carbon tetrachloride.

This means that the reaction will produce the same number of moles of $C C {l}_{4}$ as you had of $C {S}_{2}$.

Since chlorine gas is the limiting reagent, it will determine how many moles of $C {S}_{2}$ will actually react.

0.235cancel("moles"Cl_2) * ("1 mole "CS_2)/(color(red)(3)cancel("moles"Cl_2)) = "0.0783 moles"# $C {S}_{2}$

The reaction will produce 0.0783 moles of $C C {l}_{4}$. The mass of the compound that contains this many moles is

$0.0783 \cancel{\text{moles") * "153.81 g"/(1cancel("mole")) = color(green)("12.0 g } C C {l}_{4}}$