Taking into account the reaction stoichiometry, 4.57 grams of H₂O are formed when8.63 grams of Al₂O₃ reacts with HNO₃.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Al₂O₃ + 6 HNO₃ → 2 Al(NO₃)₃ + 3 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:
- Al₂O₃: 1 mole
- HNO₃: 6 moles
- Al(NO₃)₃: 2 moles
- H₂O: 3 moles
The molar mass of the compounds is:
- Al₂O₃: 102 g/mole
- HNO₃: 63 g/mole
- Al(NO₃)₃: 213 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 1 mole ×102 g/mole= 102 grams
- HNO₃: 6 moles ×63 g/mole= 378 grams
- Al(NO₃)₃: 2 moles ×213 g/mole= 426 grams
- H₂O: 3 moles ×18 g/mole= 54 grams
The following rule of three can be applied: if by reaction stoichiometry 102 grams of Al₂O₃ form 54 grams of H₂O, 8.63 grams of Al₂O₃ form how much mass of H₂O?
<u><em>mass of H₂O= 4.57 grams</em></u>
Then, 4.57 grams of H₂O are formed when8.63 grams of Al₂O₃ reacts with HNO₃.
Learn more about the reaction stoichiometry: