Respuesta :
Answer
143.49 g Al
Procedure
Using the following balanced equation determine the limiting reagent.
2AI(s) + Fe₂O₃(aq) --> AI₂O₃(aq) + 2Fe(s)
First, convert all the reagents to moles and then use the mole ratios method
[tex]150.93\text{ g Al}\frac{1\text{ mol Al}}{26.98\text{ g Al}}=5.59\text{ }mol\text{ Al}[/tex][tex]19.26\text{ g Fe}_2\text{O}_3\frac{1\text{ mol Fe}_2\text{O}_3}{159.68\text{ g Fe}_2\text{O}_3}=0.1206\text{ }mol\text{ Fe}_2\text{O}_3[/tex]Molar ratios
[tex]5.59\text{ mol Al}\frac{2\text{ mol Fe}}{2\text{ mol Al}}=5.59\text{ }mol\text{ Fe}_[/tex][tex]0.1206\text{ mol Fe}_2\text{O}_3\frac{2\text{ Fe}}{\text{ 1 mol Fe}_2\text{O}_3}=0.2412\text{ Fe}_[/tex]Therefore the limiting reagent is Fe₂O₃
Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess reagent given.
[tex]19.26\text{ g Fe}_2\text{O}_3\frac{1\text{ mol Fe}_2\text{O}_3}{159.68\text{ g Fe}_2\text{O}_3}\frac{2\text{ mol Al }}{1\text{ mol Fe}_2O_3}\frac{26.98\text{ g Al}}{1\text{ mol Al}}=6.5085\text{ g Al}[/tex]Excess = Mass of total excess reagent given – mass of excess reagent consumed in the reaction
[tex]Excess=150.93-6.5085=143.49\text{ g Al}[/tex]