Solid aluminum (ai) and oxygen (o_2) gas react to form solid aluminum oxide (al_2o_3). suppose you have 7.0 mol of al and 9.0 mol of o_2 of in a reactor. calculate the largest amount of al_2o_3 that could be produced. round your answer to the nearest 0.1 mol.

The largest amount of Al₂O₃ produced from the reaction is 3.5 moles

We'll begin by determining the limiting reactant.

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.

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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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 correct answer is:

a. formation of precipitation