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Illusion [34]
4 years ago
14

How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)?

Chemistry
1 answer:
soldier1979 [14.2K]4 years ago
7 0

Answer:

The no. of moles of CaCl₂ are needed = 3.1 mol.

Explanation:

  • From the balanced reaction:

<em>2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂,</em>

It is clear that 2 mol of AgNO₃ react with 1 mol of CaCl₂ to produce 2 mol of AgCl and 1 mol of Ca(NO₃)₂.

<u><em>Using cross multiplication:</em></u>

2 mol of AgNO₃ need  → 1 mol of CaCl₂ to react completely, from stichiometry.

6.2 mol of AgNO₃ need  → ??? mol of CaCl₂ to react completely.

∴ The no. of moles of CaCl₂ are needed = (1 mol)(6.2 mol)/(2 mol) = 3.1 mol.

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In a galvanic cell, electrons are transferred from one half cell to the other as the redox reaction progresses. what happens in
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The half cell in which the electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.

<h3><u>What is a Galvanic cell ?</u></h3>

Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.

Two half-cells make up most electrochemical cells. The half-cells allow electricity to pass via an external wire by separating the oxidation half-reaction from the reduction half-reaction.

<h3><u>Oxidation:</u></h3>

The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.

<h3><u>Reduction:</u></h3>

The second half-cell, cathode, which is frequently displayed on a figure's right side. The cathode is where reduction happens. The circuit is completed and current can flow by adding a salt bridge.

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2 years ago
A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

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