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Sidana [21]
3 years ago
15

A reaction was performed in which 0.55 g of 2‑naphthol was reacted with a slight excess of allyl bromide to make 0.59 g of allyl

2‑naphthyl ether. Calculate the theoretical yield and percent yield for this reaction. Allyl bromide has a density of 1.40 g/mL .
Chemistry
1 answer:
Ronch [10]3 years ago
6 0

Answer:

  • Theoretical yield = 0.7028 g
  • Percent yield = 83.95%

Explanation:

The reaction that takes places is:

  • C₁₀H₈O + C₃H₅Br → C₁₃H₁₂O + HBr

The<u> theoretical yield</u> is calculated when assuming that <em>all of the limiting reactant transformed into the product</em>:

  • 0.55 g naphthol ÷ 144gNaphthol/mol * \frac{1molAllyl-Naphthyl Ether}{1molNaphthol} * 184 gEther/mol = 0.7028 g

The <u>percent yield</u> is calculated using the theoretical yield and the experimental yield:

  • 0.59g / 0.7028 g * 100% = 83.95%
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M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?

Putting values in above equation, we get:

0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL

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Answer:

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Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

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% of H = 14.3

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% moles of H = 14.3 / 1.00784 = 14.19

Taking the simplest ratio for C and H as:

7.14 : 14.19 = 1 : 2

The empirical formula is = CH_2

Also, Given that:

Pressure = 508 Torr

Temperature = 17 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (17 + 273.15) K = 290.15 K  

Volume = 1.500 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.torr/K.mol

Applying the equation as:

508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K  

⇒n = 0.0421 moles

Given that :  

Amount  = 1.77 g

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0421\ moles= \frac{1.77\ g}{Molar\ mass}

Molar mass of the hydrocarbon = 42.04 g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1= 14 g/mol

Molar mass = 42.04 g/mol

So,  

Molecular mass = n × Empirical mass

42.04 = n × 14

⇒ n = 3

<u>The formula of hydrocarbon = C_3H_6</u>

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