4Al(s)+3O2(g)__> 2Al2O3(s) How many moles of Al(s) react completely with 4.50 moles of O2(g) to produce 3.00 moles of Al2O3(s)? A. 2 mol b. 1.50 mol c. 6.00 mol d. 4.00 mol show work

Option c. 6 moles of Al

Explanation:

The chemical reaction is:

4Al (s) + 3O₂ (g) → 2Al₂O₃(s)

I have 4.50 moles of oxygen. Ratio is 2:3

If 2 moles of oxide were produced by 3 moles of O₂

I made 3 moles of oxide, so I used (3 .3) / 2 = 4.5 moles of O₂

I get enough amount of oxgen, so the limiting reactant is the Al.

2 moles of oxide were produced by 4 moles of Al

So, 3 moles of oxide were produced by (3 . 4) / 2 = 6 moles of Al

If I use the reactants, I get the same value.

3 moles of oxygen needs 4 moles of Al to react

Then, 4.5 moles of oxygen must need (4.5 . 4) /3 = 6 moles of Al

There will react 6.00 moles Al and 4.50 moles O2 to produce 3.00 moles Al2O3 (option C)

Explanation:

Step 1: Data given

Number of moles O2 = 4.50 moles

Moles of Al2O3 = 3.00 moles

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)

Step 3: Calculate moles of Al

For 3.00 moles Al2O3 produced, we need 2*3.00 = 6.00 moles Al

There will react 6.00 moles Al and 4.50 moles O2 to produce 3.00 moles Al2O3 (option C)