# 2 Cl2(g) + C2H2(g) ‡ C2H2Cl4(l) How many liters of chlorine will be needed to make 75.0 grams of C2H2Cl4? Pressure: 1 atm Temperature: 298K

Feb 11, 2015

You'd need $\text{21.8 L}$ of chlorine gas to produce $\text{75 g}$ of ${C}_{2} {H}_{2} C {l}_{4}$, or 1, 1, 2, 2-tetrachloroethane, under those conditions for pressure and temperature.

You've got your balanced chemical equation

${C}_{2} {H}_{2 \left(g\right)} + 2 C {l}_{\left(g\right)} \to {C}_{2} {H}_{2} C {l}_{4 \left(l\right)}$

Every time you must decide how much of something is needed to produce a certain amount of something else, you must look at the mole ratio between the two compounds.

In your case, 2 moles of chlorine gas are needed to produce 1 mole of tetrachloroethane, which implies that there's a $\text{2:1}$ mole ratio between the two.

Use the mass of tetrachloroethane to determine how many moles were produced

$\text{75.0 g" * ("1 mole")/("168 g") = "0.446 moles}$ ${C}_{2} {H}_{2} C {l}_{4}$

The mole ratio then tells you that you need twice as many moles of chlorine for the reaction to produce that much product

$\text{0.446 moles "C_2H_2Cl_4 * ("2 moles "Cl_2)/("1 mole " C_2H_2Cl_4) = "0.892 moles}$ $C {l}_{2}$

You can determine the volume by using the ideal gas law equation, $P V = n R T$.

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

V = ("0.892 moles" * 0.082("L" * "atm")/("mol" * "K") * "298 K")/("1 atm") = "21.8 L"