Question #226981

Hydrogen bromide with a volume of 224 liters (n. o.) reacted with an excess of ethylene. The mass of the formed bromoethane

is equal to


Expert's answer

CH2+HBr→CH3CH2Br

Mole ratio = 1:1

Molar mass of hydrogen bromide = 80.91

= 224000/80.91

= 2768.51 moles

Molar mass of bromoethane = 108.97

= 2768.51 × 108.97

= 301684.34g


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