(Answered) 11 g of sodium bromide nabr is mixed with 9 g of butanol

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11 g of sodium bromide NaBr is mixed with 9 g of butanol CH3CH2CH2CH2OH to yield 258 mg of bromobutane CH3CH2CH2CH2Br. Under the catalysis of 10 ml of sulfuric acid density 1.84g/ml.

a) Write the chemical equation of the reaction
b) Find the number of mole of all reactant involved in the reaction.

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