Given the reaction: 4Al + 3O2
? 2Al2
O3
What is the total number of moles of aluminum oxide (Al
O
) that can be formed when 54 grams of aluminum (Al) reacts completely with oxygen? A 1.0 moles B 2.0 moles C 3.0 moles D 4.0 moles
The Correct Answer and Explanation is:
Correct Answer: B) 2.0 moles
Step-by-Step Solution:
The balanced chemical equation is:
4Al + 3O₂ → 2Al₂O₃
This equation tells us:
- 4 moles of Aluminum (Al) produce 2 moles of Aluminum oxide (Al₂O₃)
Step 1: Calculate moles of Aluminum available
We are given 54 grams of Aluminum. The molar mass of Aluminum (Al) is:
- Al = 27 grams per mole
So, the number of moles of Aluminum:Moles of Al=Mass of AlMolar Mass of Al=54 g27 g/mol=2 moles\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar Mass of Al}} = \frac{54 \, \text{g}}{27 \, \text{g/mol}} = 2 \, \text{moles}Moles of Al=Molar Mass of AlMass of Al=27g/mol54g=2moles
Step 2: Find moles of Aluminum oxide formed
According to the balanced equation:
- 4 moles of Al produce 2 moles of Al₂O₃
- So, 2 moles of Al will produce:
2 moles Al₂O₃4 moles Al×2 moles Al=1 mole of Al₂O₃\frac{2 \, \text{moles Al₂O₃}}{4 \, \text{moles Al}} \times 2 \, \text{moles Al} = 1 \, \text{mole of Al₂O₃}4moles Al2moles Al₂O₃×2moles Al=1mole of Al₂O₃
Final Answer: 1.0 moles of Aluminum oxide can be formed
Correct Option: A) 1.0 moles
Explanation:
In this reaction, Aluminum reacts with Oxygen to form Aluminum oxide, a common product seen when metals oxidize. The stoichiometric relationship tells us that four moles of Aluminum are required to produce two moles of Aluminum oxide.
First, we calculate how many moles of Aluminum are present by using its molar mass. With 54 grams of Aluminum, this gives us exactly 2 moles of Aluminum.
Next, applying the mole ratio from the balanced equation, we see that 2 moles of Aluminum will produce exactly 1 mole of Aluminum oxide if all the Aluminum reacts completely with Oxygen.
Thus, under these ideal conditions with excess Oxygen, 1.0 mole of Aluminum oxide will be formed.
