# You have 75.0 grams of C2H2Cl4, how many L of Cl2 do you have?

Apr 3, 2014

So, you're given 75.0 grams of ${C}_{2} {H}_{2} C {l}_{4}$ and you want to find how many L of $C {l}_{2}$.

Reaction: $2 C {l}_{2}$ + ${C}_{2} {H}_{2}$ --> ${C}_{2} {H}_{2} C {l}_{4}$

Understand the information you need to fill in your conversion factors.
You'll want to know:

1. How many grams are in 1 mol C2H2Cl4 (167.84 g) [added atomic weight from periodic table]
2. Mol to mol ratio of Cl2 and C2H2Cl4 (2:1) [given in reaction]
3. How many liters are in a mole (22.4 L/mol) [universal]

Then, you want to set up your conversion factor.
$75.0 g {C}_{2} {H}_{2} C {l}_{4}$ x $\left(\text{1 mol "C_2H_2Cl_4)/("167.84g } {C}_{2} {H}_{2} C {l}_{4}\right)$ x $\left(\text{2 mol "Cl_2)/("1 mol } {C}_{2} {H}_{2} C {l}_{4}\right)$ x $\frac{22.4 L}{1 m o l}$= ?