Question

3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O
If you begin a reaction with 4.690 g of nitric acid, how many grams of copper (II) nitrate can you theoretically produce, assuming an excess of solid copper is present?

Answer 1

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.

Reaction stoichiometry

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

Mass of Cu(NO₃)₂ produced

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?

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

mass of Cu(NO₃)₂=  5.2634 grams

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:

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