4Al + 3O2 -> 2Al2O3 Which reactant is limiting if 0.32 mol Al and 0.26 mol O2 are available?

How many moles of Al2O3 are formed from the reaction of 6.38 x 10^-3 mol of O2 and 9.15 x 10^-3 mol of Al?

If 3.17 g of Al and 2.55 g of O2 are available, which reactant is limiting?

Substances are consumed according to the coefficients in front of them.

0.32 mol of Al is consumed, so that means we use 0.08 mol per coefficient of one in front of the substances. Thus, we use 0.3 × 0.08 = 0.24 mol of O₂ gase. We do not use all of the O₂ gase, but in contrast, we use all of the Al. Hence, Al is the limiting reactant here. I personally divide the number of moles by the coefficient of the reactant and whichever of them is smaller, it's the limiting reactant.

Limiting reactant may change depending on how many moles we take from both reactants. Here, we have to determine which one is the limiting one. For simplification, we can ignore 10⁻³s.

For O₂: 6.38 ÷ 3 ≅ 2.12

For Al: 9.15 ÷ 4 ≅ 2.28

O₂ is the limiting reactant.

Reaction calculations are done according to the limiting reactants' being consumed. If 9.15 × 10⁻³ mole of O₂ is consumed and, of course its coefficient is 3, the amount of Al₂O₃, whose coefficient is 2, will be (2 ÷ 3) × 9.15 × 10⁻³ = 6.1 × 10⁻³ mol

We are given the masses of the substances. We have to find, firstly, their number of moles. Molar mass of Al is 27 gr. while that of O₂ is 18.

3.17 ÷ 27 will give us the number of moles of Al, which equals ≅ 0.12

2.55 ÷ 18 will give us the number of moles of O₂, which equals ≅ 0.14

We will divide the number of moles by the coefficients for both substances and the smaller one will be the limiting reactant.

For Al, ≈0.12 ÷ 4 ≅ 0.03

For O₂, ≈0.14 ÷ 3 ≅ 0.05

The smaller one is that of Al, so Al is the limiting reactant.

If there is anything that you have a difficulty understanding or you think anything is wrong, please let me know, everyone makes mistakes and I am sorry for my bad English ^^