Methyl salicylate, C8H8O3, is prepared by heating salicylic acid, C7H603, with
methanol, CH3OH, according to the reaction C7H603 + CH3OH ---> C8H803 +
H20 In an experiment, 2.1 g of salicylic acid is reacted with 13.20 g of methanol.
Calculate the amount, in grams, of water produced in this experiment.
Answers: 2
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Methyl salicylate, C8H8O3, is prepared by heating salicylic acid, C7H603, with
methanol, CH3OH, acc...
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