A Moving to another question wil save this response. 17 19 Question 17 3 points

Question

A Moving to another question wil save this response. 17 19 Question 17 3 points Save Answer Consider the reaction of naphthol with iodopropane to form propoxynaphthalene, shown below. NaOH, H2O CHOH mw 169.99 g/mol mw 144 g/mol mw 186.25 g/mol d 1.743 ghmL mp 122 C You need to synthesize 1 gram of propoxynaphthalene. Assuming that naphtholis the limiting reagent how much naphthol will you need? (white your answer in the form xJoox Question 17 of 19 A Moving to another question will save this response.

Explanation / Answer

1) If we see the raection, the ratio of naphthol : propyl iodide is 1:1. to give 1 mole of propoxynaphthalene.

That is one mole of naphthol gives one mole of propoxynaphthalene.

Step 1) We need to calculate moles of propoxynaphthalene -

So, number of moles = Mass in gram / Molar mass

= 1 g / 186.25 g/mol

= 0.005369 moles of propoxynaphthalene

So, one mole of naphthol gives one mole of propoxynaphthalene,, therefore

0.005369 moles are prepared from 0.005369 moles of naphthol as ratio 1:1

step 2- Moles of naphthol are converted to mass of naphthol :

Moles = Mass / Molar mass

mass = Moles x Molar mass

= 0.005369 x 144 g/mol

Mass = 0.773g og naphthol.

So,For preparation og 1 g of propoxynaphthalene, we need 0.773 g of naphthol.

2) 5 equivalent means 5 moles.

Step 1 ) We need to convert moles to mass =

Mass = Moles x Molar mass

= 5 x 169.99 g/mol

= 849.95 g

Now grams ae converted to mL using density,

Density = Mass / volume

Volume = Mass / Density

= 849.95 g / 1.743 g/mL

Volume = 487.63 mL

= 4.87 x 10-2 mL

3) It used as catalyst. It speed up the reaction by acting as phase transfer catalyst and recoverd at the end of reactoin.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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