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Answer to Question #226981 in Inorganic Chemistry for Gwaivu Yahaya
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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