What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al (molar mass 26.98 g mol-1) and 19.0 grams of O2 (molar mass 32.00 g mol-1)? Equation: 4Al + 3O2 → 2Al2O3

Al is a limiting reagent. A further explanation is below.

Given equation,

So,

4 moles of Al reacts with 3 moles of O₂ to produce 2 moles of Al₂O₃.

then,

Number of moles of Al,

=

=

Now,

then,

Number of moles of O₂,

=

=

Now,

of O₂ can react with,

=

=

But there is only moles of Al available.

Thus the above approach is right.

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Al

Explanation:

4 Al  +  3 O₂  →  2 Al₂O₃

You need to figure out which one has the smaller mole ratio.  Convert both substances from grams to moles.

(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al

(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂

Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.

(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃

(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃

Since aluminum produces the least amount of product, this is the limiting reagent.