Question Not yet answered Marked out of 1.00000 Use the following atomic weights

Question

Question Not yet answered Marked out of 1.00000 Use the following atomic weights and quantities to calculate the overall % yield of (E)-stilbene. Remember that your yeld 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, H = 1, O = 16, P = 31 , Br = 80 Flag question henylphosphonium +benzaldehyde stilbene bromide formula formula weight density grams mL moles 1.044 16.850 6.374 4.439

Explanation / Answer

1 mole of benzyltriphenylphosphonium bromide + 1 mole of benzaldehyde -----> 1 mole of stilbene

molecular weight of benzyltriphenylphosphonium bromide = 433.32

molecular weight of benzaldehyde= 106.12

molecular weight of stilbene= 180.25

Thus 433.32 g of benzyltriphenylphosphonium bromide + 106.12 g of benzaldehyde gives 180.25 g of stilbene.

Density of benzaldehyde = 1.04 g/mL

Mass of benzaldehyde = Density*volume = 1.044*4.439 =4.634 g

Mass of benzyltriphenylphosphonium bromide = 16.850 g

16.850 g of benzyltriphenylphosphonium bromide reacts with (106.12*16.850)/433.32 =4.126 g of benzaldehyde to produce (180.25*16.850)/433.32 = 7.009 g of stilbene.

Moles of benzyltriphenylphosphonium bromide = 16.850 g/433.32 g= 38.885 *10-3 moles.

As the stoichiometry of the reaction is 1:1 giving 1 mole of product, number of moles of benzaldehyde reacting is also 38.885 *10-3 moles and number of moles of stilbene as product is also 38.885 *10-3 moles.

% yiled of the product = experimental mass* 100/theoretical mass

= 6.374/7.009

=90.94%, rounding off to 91%

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.