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First we need to balance the given chemical reaction
4Al + 3O2 ----> 2Al2O3.
Second divide the given moles with stoichiometric coefficient form the reaction.
0.32 moles of Al.
0.26 moles of O2.
For Al = 0.32/4 = 0.08.
For O2 = 0.26/3 = 0.086
Now find the least number among the calculated value and that will be the limiting reactant.
They are almost the same but O2 has slightly greater value that is 0.086 than Al.
So Al is considered as the limiting reactant.
Here we need to balanced the reaction first,
4 Al +3 O2 → 2 Al2O3
imiting reactant are the one which is completely utilised in the reaction.
The molar ratio is4 Al : 3 O2 . That is 4 mole of Al combines with 3 mole of O2.
we have 0.32 mol Al,
0.32 mol Al * (3 mol O2/ 4 mol Al)
0.24 mol O2
So for 0.32 mol Al requries 0.24 mol O2, but we have 0.26 mol O2.
So Al is used completely. So Al is the limiting reagnet.
The limiting reactant in a chemical equation is the reactant that is completely consumed when the chemical reaction takes place. As a result the reactant determines for how long the reaction takes place.
The reaction given is that between aluminum and oxygen to yield Al2O3. The equation of this chemical reaction is:
4Al + 3O2 --> 2Al2O3
0.32 moles of Al and 0.26 moles of O2 are available in the reaction.
The ratio of the number of moles of Al and O2 is 0.32/0.26 = 1.23 but the reaction requires their ratio to be 4/3 = 1.33
The amount of aluminum available is less than what is required for both the reactants to be completely consumed.
Therefore aluminum is the limiting reactant and when the reaction ends oxygen is left.
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