Question 1 Not yet answered Marked out of Use the following atomic weights and q

Question

Question 1 Not yet answered Marked out of Use the following atomic weights and quantities to calculate the overall % yield of (E)-stilbene. Remember that your yield must be based on the limiting reagent, volumes must be converted to grams, and grams to moles. Filling out most of the chart will help. Give only two significant digits in your answer. If after rounding the answer is a whole number, do not include a decimal point. C . 12. ?-1,0-16, ?-31, Bra 80 100000 P Flag question benzyltriphenylphosphonium +benzaldehyde + benza bromide formula formula weight density grams mL. moles 044 32.978 3.427 1.973 yield

Explanation / Answer

Benzyltriphenylphosphonium bromide

Formula - C25H22BrP

Formula weight - 433 g/mol

Benzaldehyde

Formula - C6H5CHO

Formula weight - 106 g/mol

E-stillbene

Formula - C6H5CH=CHC6H5 or C14H12

Formula weight - 180 g/mol

Moles of C25H22BrP = mass/molecular weight

= 32.978/433

= 0.076162 mol

Mass of Benzaldehyde = volume x density

= 1.973 mL x 1.044 g/mL

= 2.059812 g

Moles of Benzaldehyde = mass/molecular weight

= 2.059812 g / 106 g/mol

= 0.019432 mol

Moles of E-stillbene = mass/molecular weight

= 3.427g / 180 g/mol

= 0.01904 mol

The reaction is

C25H22BrP + C6H5CHO = C14H12

From the stoichiometry of the reaction

1 mol C25H22BrP required = 1 mol C6H5CHO

0.076162 mol C25H22BrP required = 0.076162 mol C6H5CHO

But we have only 0.019432 mol which is less than required.

Limiting reactant = C6H5CHO

Moles of E-stillbene produced = moles of C6H5CHO consumed

= 0.019432 mol

Theoretical yield of E-stillbene produced

= 0.019432 mol x 180 g/mol

= 3.49776 g

% yield of E-stillbene = actual yield x 100/theoretical yield

= 3.427 x 100 / 3.49776

= 97.98 %

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