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Furkat [3]
3 years ago
9

Wine goes bad soon after opening because the ethanol ch3ch2oh in it reacts with oxygen gas o2 from the air to form water h2o and

acetic acid ch3cooh , the main ingredient of vinegar. what mass of ethanol is consumed by the reaction of 3.84g of oxygen gas? be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
lilavasa [31]3 years ago
8 0
The balanced chemical equation of the reaction described above is,

                            C2H6O + O2 --> H2O  + C2H4O2

If we have 3.84 g of oxygen, we divide by its molar mass.
 
                               n = (3.54 g Oxygen gas) x (1 mole O2/ 32 g O2)
                                 n = 0.11 moles O2

Using ratio and proportion,

               number of moles of ethanol = (0.11 moles O2) x (1 mole C2H6)
                                                  = 0.11 moles C2H6

Then, we multiply the calculated value to its molar mass, 46 grams /mol.
                    mass of ethanol = (0.11 mol) x (46 grams / mol)
                                                = <em>5.06 grams</em>
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Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

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(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

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Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

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