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

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

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Answer 1 · 2014-04-03T15:32:41

So, you're given 75.0 grams of $C_{2} H_{2} C l_{4}$ $C_2H_2Cl_4$ and you want to find how many L of $C l_{2}$ $Cl_2$ .

Reaction: $2 C l_{2}$ $2Cl_2$ + $C_{2} H_{2}$ $C_2H_2$ --> $C_{2} H_{2} C l_{4}$ $C_2H_2Cl_4$

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

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

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

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

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