Question

Scenario A

Wilma was given the task to produce 78 grams of aluminum oxide, Al2O3. She had one tank of oxygen gas, O2 and 45 grams of solid aluminum, Al. She knew that if she heated the aluminum in the presence of the oxygen gas, she could produce the aluminum oxide. Since she had such little amount of aluminum, she decided to use as little of aluminum as she could to produce the 78 grams of aluminum oxide. She measured out 41.3 grams of aluminum and heated it with excess oxygen. Assume Wilma’s reaction went 100%. Wilma was very happy that her reaction produced the 78 grams.
Scenario B
Fred was given the task to produce 63 grams of aluminum bromide, AlBr3. He had a tank of bromine gas, Br2, and 3 grams of solid aluminum, Al. He knew that if he heated the aluminum in the presence of excess bromine gas, he could produce the aluminum bromide. Since he had such little amount of aluminum, he decided to use as little of aluminum as he could to produce the 63 grams of aluminum bromide. He measured out 2.89 grams of aluminum and heated it with excess bromine gas. Assume Fred’s reaction went 100%. Fred was very happy his reaction produced the 63 grams.

Answer 1

To find the number of moles of Al₂O₃ produced, convert the mass of H₂O to moles and use the stoichiometric ratio from the balanced equation.

Step-by-step explanation:

In order to find the number of moles of Al₂O₃ produced when 23.9 g of H₂O are reacted according to the chemical equation 2 AlCl₃ + 3 H₂O(l) → Al₂O₃ + 6 HCl(g), we need to use stoichiometry.

First, we convert the mass of H₂O to moles by dividing it by the molar mass of H₂O (18.015 g/mol).

Once we have the moles of H₂O, we can use the stoichiometric ratio from the balanced equation to determine the moles of Al₂O₃ produced. Since the ratio is 2:1, we multiply the moles of H₂O by the ratio to find the moles of Al₂O₃.