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daser333 [38]
2 years ago
5

4. a compound called pyrene has the empirical formula c8h5. when 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boilin

g point of the solution is 85.1°c calculate the molar mass of pyrene and determine its molecular formula. the molal boiling-point constant for benzene is 2.53°c/m. its normal boiling point is 80.1°c
Chemistry
1 answer:
e-lub [12.9K]2 years ago
3 0

The molecular mass of pyrene is 204.4 g/mol.

From;

ΔT = Kb m i

Where;

  • ΔT = boiling point elevation
  • Kb = boiling point constant
  • m = molality
  • i = Van't Hoff factor

Since the compound is molecular; i = 1

The number of moles of pyrene = 4.04 g/MM

Where; MM = molar mass of pyrene

molality = number of moles of pyrene/mass of solvent in Kg

The mass of solvent = 10 g or 0.01 Kg

molality =  4.04 g/MM/0.01

ΔT = Boiling point of solution - Boiling point of pure solvent

ΔT = 85.1°C - 80.1°C

ΔT = 5°C

5 = 2.53 × 4.04 g/MM/0.01 × 1

5 = 10.22 × 1/0.01 MM

0.05MM = 10.22

MM= 10.22/0.05

MM= 204.4 g/mol

Learn more: brainly.com/question/2292439

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Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

<u>Step 1: </u>Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

<u>Step 2:</u> The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

<u>Step 3:</u> Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

<u>Step 5: </u>Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

<u>Step 8:</u> Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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This isn’t anything related to chemistry dude
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