Methanol (CH3OH) is converted to Bromomethane (CH3Br) as follows: CH3OH +HBr= CH3Br+H2O If 8.01 g of...

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Question

Methanol (CH3OH) is converted to Bromomethane (CH3Br) as follows:

CH3OH +HBr= CH3Br+H2O

If 8.01 g of Bromomethane are produced when 4.051 g of methanol is reacted with excess HBr, what is the percentage yield?

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