If 8.63 grams of Aluminum oxide react with Nitric acid, how many grams of water will be produced?

1 answer:

8 0

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

<h3>Mass of water formed</h3>

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?

$mass H_{2} O=\frac{8.63 grams of Al_{2} O_{3} x54 grams of H_{2} O}{102 grams of Al_{2} O_{3}}$