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Oduvanchick [21]
3 years ago
7

`Let's just consider the concept of stoichiometry without any confusing chemicals. Here is a generic equation: 2A + 6B ? 3C. If

you mix 4 units of A and 11 units of B, what would be the theoretical yield of C?
A. 8

B. 3

C. 6

D. 5.5

E. none of the above
Chemistry
1 answer:
lesya [120]3 years ago
7 0

Answer:

Option D. 5.5

Explanation:

The equation is this:

2A + 6B  ⇒  3C

With the amounts that we were given, let's determine which is the <em>limting reactant</em>

2 A reacts with 6 B

4 A will react with ( 4 .6)/2 = 12B

I have 11 B, so the limiting is B

6 B react with 2 A

11 B will react with (11 .2 )/6 =3.66 A

I have 4 A, so A is the excess.

6 B produce 3 C

11 B will produce ( 11 .3)/6 = 5.5C

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The tarnish that forms on objects made of silver is solid silver sulphide. This can be removed by reacting it with aluminium met
steposvetlana [31]

Answer:

5.83 mol.

Explanation:

  • From the balanced reaction:

<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>

It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.

Al reacts with Ag₂S with (2: 3) molar ratio.

<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>

<em />

  • The limiting reactant is Ag₂S.
  • The excess "left over" reactant is Al.

The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.

6 0
3 years ago
Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

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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