2 Benzaldehyde + Acetone ----NaOH, EtOH/H2O-------> Dibenzalacetone + 2H2O Place

Question

2 Benzaldehyde +  Acetone  ----NaOH, EtOH/H2O------->   Dibenzalacetone + 2H2O Place 0.50g NaOH, 5.0 mL water, and 4.0 mL ethanol in 25 mLErlenmyer flask. Place 0.50 mL benzaldehyde and theoreticalquantity acetone in a 10 mL Erlenmyer flask......... How do I find the theoretical quanitity in mL of acetoneneeded to run this reaction? When I'm figuring out the theoretical yield of benzaldehyde, Iget to 0.004947 moles benzaldehyde from 0.50 mL * 1.05 g/mL *1 mol/106.12 g benzaldehyde. Do I use the 2 moles in theoriginal equation here or do I use it to figure out the amountacetone? If anyone could walk me through the steps of figuring thisout, I'd greatly appreciate it and will rate lifesaver! :) Thank you! 2 Benzaldehyde +  Acetone  ----NaOH, EtOH/H2O------->   Dibenzalacetone + 2H2O Place 0.50g NaOH, 5.0 mL water, and 4.0 mL ethanol in 25 mLErlenmyer flask. Place 0.50 mL benzaldehyde and theoreticalquantity acetone in a 10 mL Erlenmyer flask......... How do I find the theoretical quanitity in mL of acetoneneeded to run this reaction? When I'm figuring out the theoretical yield of benzaldehyde, Iget to 0.004947 moles benzaldehyde from 0.50 mL * 1.05 g/mL *1 mol/106.12 g benzaldehyde. Do I use the 2 moles in theoriginal equation here or do I use it to figure out the amountacetone? If anyone could walk me through the steps of figuring thisout, I'd greatly appreciate it and will rate lifesaver! :) Thank you!

Explanation / Answer

Moles of benzaldehyde = 0.50 mL * 1.05 g/mL * 1mol / 106.12 gbenzaldehyde                                     =0.00494 moles . From the equation, we know 2 moles of benzaldehyde react withone mole of acetone. So theoretical amount of acetone = 0.00494 / 2                                                  = 0.00247 moles . Theoretical volume of acetone = 0.00247 moles * (58.08 g / 1mole) / (0.79 g/ mL)                                              = 0.182 mL

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