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kondor19780726 [428]
2 years ago
13

3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O

Chemistry
1 answer:
AysviL [449]2 years ago
7 0

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 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:

  • Cu: 3 moles
  • HNO₃: 8 moles
  • Cu(NO₃)₂: 3 mole
  • NO: 2 moles
  • H₂O: 4 moles

The molar mass of the compounds is:

  • Cu: 63.55 g/mole
  • HNO₃: 63 g/mole
  • Cu(NO₃)₂: 187.55 g/mole
  • NO: 30 g/mole
  • H₂O: 18 g/mole

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

  • Cu: 3 moles ×63.55 g/mole= 190.65 grams
  • HNO₃: 8 moles ×63 g/mole= 504 grams
  • Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
  • NO: 2 moles ×30 g/mole= 60 grams
  • H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3}  x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }

<u><em>mass of Cu(NO₃)₂=  5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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Answer:

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Explanation:

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3 years ago
How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr
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Answer:

percentage yield = 67%

Explanation:

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Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

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Number of moles = 0.08 mol

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Number of moles = mass/ molar mass

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Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

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