hi, i'm writting a aldol wittig reaction lab and am having trouble calculating my theoretical yield
the reaction is C9H10O3 + Ccc9H8O ---NaOH ----- C18H16O3
which is 3,4 dimethoxybenzaldehyde + 1-Indanone
I need this to find the limiting product so i can do basic stoich
any help appreciated
if anyone is able to do the theoretical yield calculation that would be great, I used 0.25g of C9H10O3
0.20g of 1-indanone
and 0.05 g of naoh
thanks
balancing this equation
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Start with calculations of molecular weight:
C9H10O3 FW: 166.17
1-Indanone. alpha-Hydrindone. C9H8O F.W. 132.16
NaOH FW 40
Now canlculate number of mols of each component:
C9H10O3 0.25/166.17 = 0.0015 mol
1-Indanone. 0.2/132.16 = 0.0015 mol
NaOH 0.05/40 = 0.00125 mol
So, the reagent are equimolar and NaOH is used in less amount. Is NaOH a limiting reagent? Not necessarily. Leant about the reaction and find out if NaOH is a catalyst. In that case your reaction doesn't have any limiting reagents and you should theoretically obtain 0.0015 mol of the product
C9H10O3 FW: 166.17
1-Indanone. alpha-Hydrindone. C9H8O F.W. 132.16
NaOH FW 40
Now canlculate number of mols of each component:
C9H10O3 0.25/166.17 = 0.0015 mol
1-Indanone. 0.2/132.16 = 0.0015 mol
NaOH 0.05/40 = 0.00125 mol
So, the reagent are equimolar and NaOH is used in less amount. Is NaOH a limiting reagent? Not necessarily. Leant about the reaction and find out if NaOH is a catalyst. In that case your reaction doesn't have any limiting reagents and you should theoretically obtain 0.0015 mol of the product
Remember safety first! Check MSDS and consult with professionals before performing risky experiments.