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Advocard [28]
3 years ago
8

These equations below show two standard reduction potentials. Fe3+(aq) + e– Fe2+(aq) Eo = +0.77 Cu2+(aq) + 2e– Cu(s) Eo = +0.34

Which value represents the standard cell potential of the spontaneous overall reaction? Use E cell = E red, cathode – E red, anode. –1.11 V –0.43 V +0.43 V +1.11 V
Chemistry
1 answer:
IgorC [24]3 years ago
6 0
<h3>Answer:</h3>

= +0.43 V

<h3>Explanation:</h3>

The two standard reduction potentials are;

Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E₀ = +0.77 V

Cu²⁺(aq) + 2e⁻ → Cu(s)    E₀ = +0.34 V

The overall reaction is;

Fe³⁺(aq) + Cu(s) → Fe²⁺(aq) + Cu²⁺(aq)

To calculate the standard cell potential of the overall reaction, we use the equation;

Ecell = E redu, cathode - E red, anode

From the reaction; Copper lost electrons to form copper ions (Cu²⁺) (Oxidation) while Fe³⁺ gained electrons to form Fe²⁺ (reduction)

Therefore;

E cell = + 0.77 v - (+0.34 v)

        = + 0.43 V

Hence, the standard cell potential of the overall reaction is +0.43 V

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Calculate the maximum volume in ml of 0.15M HCl that each of the following antacid formulations would be expected to neutralize.
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a. 34 mL; b. 110 mL

a. A tablet containing 150 Mg(OH)₂


Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O


<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂

= 2.572 mmol Mg(OH)₂


<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]

= 5.144 mmol HCl


Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl


b. A tablet containing 850 mg CaCO₃


CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O


<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃

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<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]

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if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?
vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 2 moles
  • H₂SO₄: 3 moles
  • Al₂(SO₄)₃. 1 mole
  • H₂: 3 moles

The molar mass of the compounds is:

  • Al: 27 g/mole
  • H₂SO₄: 98 g/mole
  • Al₂(SO₄)₃: 342 g/mole
  • H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al: 2 moles ×27 g/mole= 54 grams
  • H₂SO₄: 3 moles ×98 g/mole= 294 grams
  • Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
  • H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield} x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know:

  • actual yield= 75.26 grams
  • theorical yield= 88.67 grams

Replacing in the definition of percent yields:

percent yield=\frac{75.26 grams}{88.67 grams} x100

Solving:

<u><em>percent yield= 84.88%</em></u>

Finally, the percent yield for the reaction is 84.88%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

#SPJ1

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Explanation:

hope this helps

pls mark me brainliest

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