Question

How many liters of chlorine at 101.325 kPa and 273K will be needed to make 75.0 grams of C2H2Cl4

Answer 1

Using the ideal gas equation,

$PV= nRT$

$101325 \times V = \dfrac{75}{167.84}\times 8.31 \times 273$

$V = \dfrac{17006388}{170147.25}$

$V = 99.95L$