3 Cu + SHNO3 — 3 Cu(NO3)2 + 2 NO + 4 H20
1 answer:
Considering the reaction stoichiometry, the mass of H₂O that is produced when 11.9 moles of HNO₃ react is 107.1 grams.
<h3>Reaction stoichiometry</h3>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 moles
- NO: 2 moles
- H₂O: 4 moles
The molar mass of the compounds present in the reaction is:
- Cu: 63.54 g/mole
- HNO₃: 63 g/mole
- Cu(NO₃)₂: 187.54 g/mole
- NO: 30 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:
Cu: 3 moles ×63.54 g/mole= 190.62 grams
HNO₃: 8 moles ×63 g/mole= 504 grams
Cu(NO₃)₂: 3 moles ×187.54 g/mole= 562.62 grams
NO: 2 moles ×30 g/mole= 60 grams
H₂O: 4 moles ×18 g/mole= 72 grams
<h3>Mass of H₂O produced</h3>It is possible to determine the the amount of mass of H₂O produced by a rule of three: if by stoichiometry 8 moles of HNO₃ produce 72 grams of H₂O, if 11.9 moles of HNO₃ react how much mass of H₂O will be formed?
<u><em>mass of H₂O= 107.1 grams</em></u>
In summary, the mass of H₂O that is produced when 11.9 moles of HNO₃ react is 107.1 grams.
Learn more about reaction stoichiometry:
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Answer:
The enthalpy for given reaction is 232 kilo Joules.
Explanation:
...[1]
..[2]
..[3]
..[4]
2 × [2] + [3] - [1] ( Using Hess's law)
The enthalpy for given reaction is 232 kilo Joules.
Answer:
The value becomes 0.0821 L-atm/K⁻¹-mol⁻¹.
Explanation:
The value of gas constant is 0.0821.
We know that, the ideal gas law is as follows :
Where
P is pressure
V is volume
T is temperature
In order to use the Ideal Gas Constant of 0.0821, the units are follows :
Volume = Litre (L)
Pressure = atm
Amount = mol⁻¹
Temperature = K⁻¹
So, the value becomes 0.0821 L-atm/K⁻¹-mol⁻¹.