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Alexxx [7]
3 years ago
9

Consider the reaction below. 2H2 O2 Right arrow. 2H2O How many moles of water are produced from 13. 35 mol of oxygen? 6. 675 mol

26. 70 mol 53. 40 mol 66. 75 mol.
Chemistry
1 answer:
RUDIKE [14]3 years ago
4 0

Considering the reaction stoichiometry, the correct answer is second option: the total number of moles of water produced from 13. 35 mol of oxygen is 26.7 moles.

The balanced reaction is:

2 H₂ + O₂ → 2 H₂O

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:

  • H₂: 2 moles
  • O₂: 1 mole  
  • H₂O: 2 moles

Then you can apply the following rule of three: if by stoichiometry 1 moles of O₂ produce 2 moles of H₂O, 13.35 moles of O₂ will produce how many moles of H₂O?

amount of moles of H_{2} O=\frac{13.35 moles of O_{2} x2 moles of H_{2} O}{1 mole of O_{2}}

<u><em>amount of moles of H₂O= 26.7 moles</em></u>

Finally, the correct answer is second option: the total number of moles of water produced from 13. 35 mol of oxygen is 26.7 moles.

Learn more about stoichiometry:

  • <u>brainly.com/question/16487206?referrer=searchResults </u>
  • <u>brainly.com/question/14446695?referrer=searchResults </u>
  • <u>brainly.com/question/11564309?referrer=searchResults </u>
  • <u>brainly.com/question/4025026?referrer=searchResults </u>
  • <u>brainly.com/question/18650135?referrer=searchResults</u>
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Answer & Explanation:

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Fe^{2+} +NO_{3}^{-}  => Fe{(OH)}_3 + N_2

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺  ⇒ Fe(OH)₃

reduction

NO₃⁻  ⇒ N₂

Oxidation Half Reaction

(<em>redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)</em>

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1:   2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + <u><em>30H₂0</em></u> + 2NO₃⁻ + <u><em>12H⁺</em></u> + <u><em>10e⁻</em></u> ⇒10Fe(OH)₃ + <u><em>30H⁺</em></u><u><em> </em></u>+ <em><u>10e⁻</u></em> +              

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REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq)             E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) -  2e⁻ ⇌ N₂(g) + 18H⁺       E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂    E°cell = -2.8V

E°cell = E°red + E°ox

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