3-pentanone(C 5 H 10 O) +4-nitrobenzaldehyde(C 7 H 5 O 3 N) =aldol condensation

Question

3-pentanone(C5H10O) +4-nitrobenzaldehyde(C7H5O3N) =aldol condensation product density of 3-pentanone=0.813g/mL density of 4-nitrobenzaldehyde=1.496g/mL Calculate the quantities in milliliters of4-nitrobennzaldehyde and 3-pentanone that will need to be combinedin order to prepare 10g of the solid product 3-pentanone(C5H10O) +4-nitrobenzaldehyde(C7H5O3N) =aldol condensation product density of 3-pentanone=0.813g/mL density of 4-nitrobenzaldehyde=1.496g/mL Calculate the quantities in milliliters of4-nitrobennzaldehyde and 3-pentanone that will need to be combinedin order to prepare 10g of the solid product

Explanation / Answer

                    Given that 10 g of aldol product is obtained.            Number of moles of the solid product formed =10 g / 237.25 g/mol                                                                                       = 0.04214 mol            1mol of 3-pentanone reacts with 1 mole of 4-nitrobenzaldehydeto form one mole of solid product.                  Number of moles of 3-pentanone = 0.04214 mol                         0.04214 mol = mass of 3-pentanone / 86.13 g/mol                      mass of 3-pentanone = 3.630 g                     given density = 0.813 g/mL                     Volume of 3-pentanone taken = 3.63 g / 0.813 g/mL                                                                        = 4.465 mL             Number of moles of 4-nitro benzaldehyde = 0.04214 mol                                 0.04214 mol = mass / 151.12g/mol                                  mass of 4-nitro benzaldehyde = 6.368g                           density = 1.496 g/mL                          volume of 4-nitro benzaldehyde = 6.368 g/ 1.496 g/mL                                                                                 = 4.25 mL              Given that 10 g of aldol product is obtained.            Number of moles of the solid product formed =10 g / 237.25 g/mol                                                                                       = 0.04214 mol            1mol of 3-pentanone reacts with 1 mole of 4-nitrobenzaldehydeto form one mole of solid product.                  Number of moles of 3-pentanone = 0.04214 mol                         0.04214 mol = mass of 3-pentanone / 86.13 g/mol                      mass of 3-pentanone = 3.630 g                     given density = 0.813 g/mL                     Volume of 3-pentanone taken = 3.63 g / 0.813 g/mL                                                                        = 4.465 mL             Number of moles of 4-nitro benzaldehyde = 0.04214 mol                                 0.04214 mol = mass / 151.12g/mol                                  mass of 4-nitro benzaldehyde = 6.368g                           density = 1.496 g/mL                          volume of 4-nitro benzaldehyde = 6.368 g/ 1.496 g/mL                                                                                 = 4.25 mL

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