Answer:
8.66 g of Al₂O₃ will be produced
Explanation:
4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)
This is the reaction.
Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.
4.6 g / 26.98 g/mol = 0.170 moles
Ratio is 4:2.
4 moles of aluminum can produce 2 moles of Al₂O₃
0.170 moles of Al, may produce (0.170 .2)/ 4 = 0.085 moles
Let's convert the moles of Al₂O₃ to mass.
0.085 mol . 101.96 g/mol = 8.66 g
Answer:
There will be made 8.7 grams of Al2O3
Explanation:
Step 1: Data given
Mass of Al = 4.6 grams
Molar mass Al = 26.98 g/mol
Oxygen is in excess
Molar mass of Al2O3 = 101.96 g/mol
Step 2: The balanced equation
4Al(s) + 3O2(g) → 2Al2O3(s)
Step 3: Calculate moles Al
Moles Al = Mass Al / molar mass Al
Moles Al = 4.6 grams / 26.98 g/mol
Moles Al = 0.1705 moles
Step 4: Calculate moles Al2O3
For 4 moles Al we need 3 moles O2 to produce 2 moles of Al2O3
For 0.1705 moles Al we'll have 0.1705 / 2 = 0.08525 moles Al2O3
Step 5: Calculate mass Al2O3
Mass Al2O3 = moles Al2O3 * molar mass Al2O3
Mass Al2O3 = 0.08525 moles* 101.96 g/mol
Mass Al2O3 = 8.7 grams
There will be made 8.7 grams of Al2O3
