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2 years ago

Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:

Chemistry

1 answer:

Katarina [22]2 years ago

8 0

Answer and Explanation:

The 3 reactions represented are

C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O

C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O

C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O

ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol

ΔH°f = ΔH°(products) - ΔH°(reactants)

For reaction 1,

ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol

For reaction 2,

ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol

For reaction 3,

ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol

b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.

Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol

Number of moles = 254/296 = 0.858 mole

heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol

c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole

5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole

The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole

q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ

QED!!!

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A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.

solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al + 3Cl₂ → 2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

Cl₂ : AlCl₃

3 : 2

0.06 : 2/3×0.06 = 0.04

Al : AlCl₃

2 : 2

0.10 : 0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0

1 year ago

A solid sample of a compound and a liquid sample of the same compound are each tested for electrical conductivity. which test co

marshall27 [118]

If these are you choices:
(1) Both the solid and the liquid are good conductors.

(2) Both the solid and the liquid are poor conductors.

(3) The solid is a good conductor, and the liquid is a poor conductor.

(4) The solid is a poor conductor, and the liquid is a good conductor.

Then the answer is number 4. This is because ionic compound conducts electricity when it is dissolved in water.

8 0

2 years ago

At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)

viktelen [127]

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc. 0.500 M 0 M 0 M

At eqm. conc. (0.500-2x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x = 0.0955 M

Now put all the given values in this expression, we get :

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029

7 0

2 years ago

Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp

sergiy2304 [10]

Explanation:

The given data is as follows.

$t_{m}$ = 30.0 sec, $t_{r1}$ = 5 min = $5 \times 60 sec$ = 300 sec

$t_{r2}$ = 12.0 min = $12 \times 60 sec$ = 720 sec

Formula for adjusted retention time is as follows.

$t'_{r} = t_{r} - t_{m}$

= 300 sec - 30.0 sec

= 270 sec

$t'_{r2}$ = 720 sec - 30 sec

= 690 sec

Formula for relative retention ( $\alpha$ ) is as follows.

$\alpha = \frac{t'_{r2}}{t'_{r1}}$

= $\frac{690 sec}{270 sec}$

= 2.56

Thus, we can conclude that the relative retention is 2.56.

4 0

2 years ago

Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reacti

Degger [83]

Answer:

k = 1.3 x 10⁻³ s⁻¹

Explanation:

For a first order reaction the integrated rate law is

Ln [A]t/[A]₀ = - kt

where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.

We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus

Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt

- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹

4 0

2 years ago