Question #193559

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

Expert's answer

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"

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