Respuesta :
Answer:
There will be formed 3.5 moles of Al2O3 ( 356.9 grams)
Explanation:
Step 1: Data given
Numbers of Al = 7.0 mol
Numbers of mol O2 = O2
Molar mass of Al = 26.98 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equation
4Al(s) + 3O2(g) → 2Al2O3(s)
Step 3: Calculate limiting reactant
For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3
Al is the limiting reactant, it will be consumed completely (7 moles).
O2 is in excess. There will react 3/4 * 7 = 5.25 moles
There will remain 9-5.25 = 3.75 moles
Step 4: Calculate moles Al2O3
For 4 moles Al we'll have 2moles Al2O3
For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced
Step 5: Calculate mass of Al2O3
Mass Al2O3 = moles Al2O3 * molar mass Al2O3
Mass Al2O3 = 3.5 moles* 101.96 g/mol
Mass Al2O3 = 356.9 grams
There will be formed 3.5 moles of Al2O3 ( 356.9 grams)
The largest amount of Al₂O₃ produced from the reaction is 3.5 moles
We'll begin by determining the limiting reactant.
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore,
7 moles of Al will react with = (7 × 3)/4 = 5.25 moles of O₂
From the calculation made above, we can see that only 5.25 moles of O₂ out of 9 moles given, is needed to completely react 7 moles of Al.
Therefore, Al is the limiting reactant and O₂ is the excess reactant.
- Finally, we shall determine the largest amount of Al₂O₃ produced from the reaction. This can be obtained by using the limiting reactant as illustrated below
From the balanced equation above,
4 moles of Al reacted to produce 2 moles of Al₂O₃.
Therefore,
7 moles of Al will react to produce = (7 × 2)/4 = 3.5 moles of Al₂O₃.
Thus, the largest amount of Al₂O₃ produced from the reaction is 3.5 moles
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