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The first reaction is an esterification reaction between acetic acid (CH₃COOH) and methanol (CH₃OH). In an esterification reaction, the -OH group of the carboxylic acid reacts with the -OH group of the alcohol, forming an ester and releasing a water molecule. So in this case, the esterification of acetic acid and methanol will result in the formation of methyl acetate (CH₃COOCH₃) and a water molecule (H₂O). The balanced chemical equation for this reaction is: \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} + \mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{COOCH}_{3} + \mathrm{H}_{2}\mathrm{O} \)

The second reaction is an esterification reaction between propanol (CH₃CH₂CH₂OH) and formic acid (HCOOH). Similar to the first reaction, the esterification process involves the reaction of the -OH group of the carboxylic acid with the -OH group of the alcohol. This reaction results in the formation of an ester and a water molecule. In this case, when propanol reacts with formic acid, propyl formate (CH₃CH₂CH₂OCHO) and a water molecule (H₂O) are formed. The balanced chemical equation for this reaction is: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{HCOOH} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCHO} + \mathrm{H}_{2}\mathrm{O} \) Now we have completed both reactions: a. \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} + \mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{COOCH}_{3} + \mathrm{H}_{2}\mathrm{O} \) b. \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{HCOOH} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCHO} + \mathrm{H}_{2}\mathrm{O} \)