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Taya2010 [7]
3 years ago
13

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 3.43 g of octane is m

ixed with 3.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Lisa [10]3 years ago
8 0

Answer:

The maximum mass of carbon dioxide, CO₂ produced is 2.64 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O

Next, we shall determine the masses of C₈H₁₈ and O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C₈H₁₈ = (8×12) + (18×1)

= 96 + 18

= 114 g/mol

Mass of C₈H₁₈ from the balanced equation = 2 × 114 = 228 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 25 × 32 = 800 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 16 × 44 = 704 g

SUMMARY:

From the balanced equation above,

228 g of C₈H₁₈ reacted with 800 g of O₂ to produce 704 g of CO₂.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

228 g of C₈H₁₈ reacted with 800 g of O₂.

Therefore, 3.43 g of C₈H₁₈ will react with = (3.43 × 800)/228 = 12.04 of O₂

From the calculations made above, we can see that a higher mass (i.e 12.04 g) of O₂ than what was given (i.e 3 g) is needed to react completely with 3.43 g of C₈H₁₈. Therefore, O₂ is the limiting reactant and C₈H₁₈ is the excess reactant.

Finally, we shall determine the maximum mass of carbon dioxide, CO₂ produced from the reaction.

To obtain the maximum mass of carbon dioxide, CO₂ produced, the limiting reactant will be used because all of it is consumed in the reaction.

The limiting reactant is O₂ and the maximum mass of carbon dioxide, CO₂ produced can be obtained as follow:

From the balanced equation above,

800 g of O₂ reacted to produce 704 g of CO₂.

Therefore, 3 g of O₂ will react to produce = (3 × 704)/800 = 2.64 g of CO₂.

Thus, the maximum mass of carbon dioxide, CO₂ produced is 2.64 g

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Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

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Ka=1.53*10^-^5

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This is basics of organic chemistry - the element that has atoms that can bond with each other to form ring, chain, and network structures is (3) carbon.
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Answer:

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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
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Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

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