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8 2 H NaOH Еюн 5 benzaldehyde d 1.044 acetone d0.791 When doing part B of the reaction for lab, how many grams of benzaldehyd
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Answer #1

moles dibenzalacetone required = 15g / 234.29 g/mole = 0.06402321908 mole

1 mole dibenzalacetone is formed by 2 mole benzaldehyde, so

moles benzaldehyde required = 2 x 0.06402321908 = 0.12804643817 mole

mass of benzaldehyde required = 0.12804643817 mole x 106.12 g/mole = 13.5882880191 g

Answer: 13.588 grams

8 2 H NaOH Еюн 5 benzaldehyde d 1.044 acetone d0.791 When doing part B of the reaction for lab, how many grams of benzaldehyd

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