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)

here are some of the ways in which chemical and nuclear reactions differ.

explanation: