3 Cu + SHNO3 — 3 Cu(NO3)2 + 2 NO + 4 H20

1 answer:

3 0

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?

$mass of H_{2}O=\frac{11.9 moles of HNO_{3} x72 grams of H_{2}O}{8 moles of HNO_{3}}$