In this balanced equation 4Al + 3O2 = 2Al2O3. When 0.54 mole aluminum is reacted in a closed container with 7.80 g oxygen, how many grams of aluminum oxide can form. Please show steps to work out. Thanks
This is a limiting reagent problem. See your post above with Al and oxygen. Use the same process.
To find out how many grams of aluminum oxide can form, you'll need to use the molar ratios provided in the balanced equation and then convert the given quantities into moles. Here's how you can solve this problem step by step:
Step 1: Find the molar mass of each element.
The molar mass of aluminum (Al) is 26.98 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
Step 2: Convert the given quantity of aluminum into moles.
Given: 0.54 moles of aluminum
Step 3: Convert the given quantity of oxygen into moles.
Given: 7.80 g of oxygen
To convert grams of oxygen to moles, use the molar mass of oxygen:
7.80 g O₂ × (1 mol O₂ / 32.00 g O₂) = 0.24375 moles of oxygen
Step 4: Determine the limiting reactant.
Compare the moles of aluminum and moles of oxygen. The reactant that is completely used up will limit the amount of product (aluminum oxide) that can be formed. The balanced equation indicates that the stoichiometric ratio of aluminum to oxygen is 4:3.
To determine the limiting reactant, calculate the moles of each reactant needed to react with the other:
- For aluminum, (0.54 moles Al) × (3 moles O₂ / 4 moles Al) = 0.405 moles O₂
- For oxygen, (0.24375 moles O₂) × (4 moles Al /3 moles O₂) = 0.325 moles Al
Since oxygen requires less moles than aluminum, it is the limiting reactant.
Step 5: Calculate the moles of aluminum oxide formed.
According to the balanced equation: 4Al + 3O₂ → 2Al₂O₃
From the limiting reactant, oxygen (O₂), we can deduce the number of moles of aluminum oxide (Al₂O₃) formed.
- Moles of Al₂O₃ = (0.24375 moles O₂) × (2 moles Al₂O₃ / 3 moles O₂) = 0.1625 moles Al₂O₃
Step 6: Convert moles of aluminum oxide into grams.
To convert moles of aluminum oxide to grams, use the molar mass of aluminum oxide (Al₂O₃):
0.1625 moles Al₂O₃ × (101.96 g Al₂O₃ / 1 mol Al₂O₃) = 16.58 grams of aluminum oxide
Therefore, when 0.54 mole of aluminum is reacted with 7.80 grams of oxygen, approximately 16.58 grams of aluminum oxide can form.