Question

Solid aluminum and oxygen gas react to form solid aluminum oxide. Suppose you have 3.0 mol of Al and 11.0 mol of O2 in a reactor. Suppose as much as possible of the reacts. How much will be left? Round your answer to the nearest 0.1 mol.

Answer 1

8.75 moles of oxygen will be left

The reaction is

Al(s) + O₂(g) -----> Al₂O₃(s)

On balancing the equation, we get:

4Al(s) + 3O₂(g) -----> 2Al₂O₃(s)

• If 3 mol of Al reacts, we need 3/4 x 3 = 9/4 = 2.25 mol of oxygen
• If 11 mol of oxygen reactions, we need 11/3 x 4 = 44/3 = 14.7 mol of Aluminium
• But we only have 3 mol of Al
• Aluminium is the limiting reagent and all 3 moles of Al will react

• We get 2 moles of aluminium oxide from 4 moles of aluminium.
• So from 3 moles of aluminium, we get:

                      3/4 x 2 = 6/4 = 1.5 moles of aluminium oxide

• If we use 4 moles of Al, we need 3 moles of oxygen
• So when we use 3 moles of Al,  we need

                      3/4 x 3 = 9/4 = 2.25 mol of oxygen

• We had 11 moles of oxygen
• The amount of oxygen left = 11-2.25 = 8.75 mol

8.75 moles of oxygen will be left

Learn more about limiting reagents here:

https://brainly.com/question/23661051

#SPJ10