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Methanol (CH3OH) is converted to bromomethane (CH3BR) as follows: CH3OH + HBr → CH3B + H2...

Methanol (CH3OH) is converted to bromomethane (CH3BR) as follows: CH3OH + HBr → CH3B + H2 (balanced) If 7.00 g of bromomethane (CH3Br) are produced when 2.42 g of methanol (CH3OH) is reacted with excess HBr, what is the percentage yield?
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