Question

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

Answer 1

You'd need `"21.8 L"` of chlorine gas to produce `"75 g"` of `C_2H_2Cl_4`, or 1, 1, 2, 2-tetrachloroethane, under those conditions for pressure and temperature.

You've got your balanced chemical equation

`C_2H_(2(g)) + 2Cl_((g)) -> C_2H_2Cl_(4(l))`

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 `"2:1"` mole ratio between the two.

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

`"75.0 g" * ("1 mole")/("168 g") = "0.446 moles"` `C_2H_2Cl_4`

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

`"0.446 moles "C_2H_2Cl_4 * ("2 moles "Cl_2)/("1 mole " C_2H_2Cl_4) = "0.892 moles"` `Cl_2`

You can determine the volume by using the ideal gas law equation, `PV = nRT`.

`Pv = nRT => V = (nRT)/P`

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