Answer:
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.
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,
= [tex]\frac{10 \ g}{26.98 \ g/mol}[/tex]
= [tex]3.706\times 10^{-1} \ moles[/tex]
Now,
then,
Number of moles of O₂,
= [tex]\frac{19 \ g}{32 \ g/mol}[/tex]
= [tex]5.937\times 10^{-1} \ moles[/tex]
Now,
[tex]5.937\times 10^{-1} \ moles[/tex] of O₂ can react with,
= [tex]4\times \frac{5.937\times 10^{-1}}{3}[/tex]
= [tex]7.916\times 10^{-1} \ moles \ of \ Al[/tex]
But there is only [tex]3.706\times 10^{-1}[/tex] moles of Al available.
Thus the above approach is right.
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