can someone please help me with my chemistry lab. chemistry is so extensive and i am overwhelmed. i have no idea at this point i am so confused. i will be glad to rate you if you can please help me. i have attached the lab and the sheet i need help with. please show how you solved. PLEASE HELP ME! Mass of benzaldehyde used1.1715 g0.2903 g0.291 g1.3493 gMass of acetone usedMass of filter paperMass of filter paper + precipitateNaOHC17H140(s) + 2 H20(1)2 C,HgO () + C3H6O (1)benzaldehydeFor the reaction:acetonedibenzalacetoneYour reagents are benzaldehyde and acetone. The reactionshown above is balanced. NaOH is a catalyst and is not thereforenot included when balancing the reaction. Dibenzalacetone isyour product of interest. DATA ANALYSISMass of filter paperMass of filter paper + precipitateSHOW YOUR CALCULATIONSCTIONTheoretical YieldActual Yield% RecoveredRelevance97Limiting and Excess Reagents

The balanced reaction taking place is given as, Given : Mass of C7H6O taken = 1.1715 g Mass of C3H6O taken = 0.2903 g Molar mass of C3H6O = Atomic mass of C X 3 + Atomic mass of H X 6 + Atomic mass of O = 12 X 3 + 1 X 6 + 16 = 58 g/mol Molar mass of C7H6O = Atomic mass of C X 7 + Atomic mass of H X 6 + Atomic mass of O = 12 X 7 + 1 X 6 + 16 = 106 g/mol And molar mass of C17H14O = Atomic mass of C X 17 + Atomic mass of H X 14 + Atomic mass of O = 12 X 17 + 1 X 14 + 16 = 234 g/molSince mass = moles X molar mass => 1.1715 = moles of C7H6O taken X 106 => Moles of C7H6O taken = 0.011052 mol approx. And 0.2903 = moles of C3H6O taken X 58 => Moles of C3H6O taken = 0.0050052 mol approx. From the above balanced reaction we can see that 2 moles of C7H6O is required to react with 1 mole of C3H6O Hence moles of C7H6O required = moles of C3H6O taken X 2 = 0.0050052 X 2 = 0.01001 mol approx. Since moles of C7H6O taken is more than required. Hence C7H6O is present in excess amount and the limiting reactant is C3H6O.Since theoretically all the limiting reactant will react. And from 1 mole of C3H6O, 1 mole of C17H14O is produced. Hence moles of C17H14O produced = moles of C3H6O taken = 0.0050052 mol Hence mass of C17H14O produced theoretically = moles X molar mass = 0.0050052 X 234 = 1.1712 g approx. Hence the theoretical yield of reaction = mass of C17H14O produced theoretically = 1.1712 g approx. And experimental yield of reaction = Mass of filter paper with precipitate - mass of filter paper => Experimental yield = 1.3493 - 0.291 = 1.0583 g Hence the percent yield of the reaction is given by, => Percent yield = Experimental yield X 100 Theoretical yield = 1.0583 X 1001.1712 = 90.36 % approx. Hence the % recovered = percent yield = 90.36 %.