How many moles of O2 are needed to combine with 0.288 mol Al to give Al2O3?
Write the balanced equation.
4Al + 302 >> 2Al2O3
for every 4 moles of Al, you need three moles of O2
To determine the number of moles of O2 needed to combine with 0.288 mol Al and form Al2O3, we need to use the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between aluminum (Al) and oxygen (O2) to form aluminum oxide (Al2O3) is:
4 Al + 3 O2 -> 2 Al2O3
From the balanced equation, we can see that 4 moles of Al react with 3 moles of O2 to produce 2 moles of Al2O3.
Given that we have 0.288 mol Al, we can use the stoichiometry of the balanced equation to calculate the corresponding amount of O2 needed.
Using the ratio of 4 moles Al to 3 moles O2, we can set up a proportion:
(0.288 mol Al) / (4 mol Al) = (x mol O2) / (3 mol O2)
Solving for x, we find:
x = (0.288 mol Al * 3 mol O2) / 4 mol Al = 0.216 mol O2
Therefore, 0.216 moles of O2 are needed to combine with 0.288 mol Al to give Al2O3.