A 2.92 g (27.0 mmol) sample of phenylhydrazine was dissolved in 10 mL of 95% ethanol and stirred while 3.30 g (27.0 mmol) of salicylaldehyde, dissolved in 15 mL of ethanol, was added. what is the theoretical yield in grams for salicylaldehyde phenylhydrazone?

In this case, the undergoing chemical reaction is shown on the attached picture, in which you can see that phenylhydrazine reacts with salicylaldehyde to yield salicylaldehyde phenylhydrazone.

In such a way, considering that just the moles of the reagents are important to identify the limiting reagent and subsequently the theoretical amount of salicylaldehyde phenylhydrazone, one proceeds as follows:

[tex]n_{phenylhydrazine }^{available}=27.0 mmol\ phenylhydrazine\\n_{phenylhydrazine }^{consumed}=27.0mmol\ salicylaldehyde*\frac{1mmol\ phenylhydrazine}{1mmol\ salicylaldehyde}=27.0mmol\[/tex]

Therefore, since all of the reagents are consumed, one computes the theoretical yield of the product as shown below considering its molar mass of 212.253 g/mol or 212.253mg/mmol which is the same:

[tex]n_{salicylaldehyde\ phenylhydrazone}=27.0mmol\ phenylhydrazine*\frac{1mmol\ salicylaldehyde\ phenylhydrazone}{1mmol\ phenylhydrazine}\\\\m_{salicylaldehyde\ phenylhydrazone}=27.0mmol\ salicylaldehyde\ phenylhydrazone*\frac{212.253mg\ salicylaldehyde\ phenylhydrazone}{1mmol\ salicylaldehyde\ phenylhydrazone}*\frac{1gmg\ salicylaldehyde\ phenylhydrazone}{1000mg\ salicylaldehyde\ phenylhydrazone} \\\\m_{salicylaldehyde\ phenylhydrazone}=5.73g\ salicylaldehyde\ phenylhydrazone[/tex]

Best regards.