The limiting reactant is chlorine (Cl2).
Explanation:
Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.
2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.
Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g
= 1.260 g moles of Al
Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g
= 1.10 g moles of Cl2
Dividing each reactant by it's coefficient in the balanced equation obtains:
1.260 moles Al / 2 = 0.63 g moles of Al
1.11 moles Cl2 / 3 = 0.36 g moles of Cl2
The reactant which produces a lesser amount of product is called as limiting reactant.
Here the Limiting reactant is Cl2.
The reaction of chlorine and aluminum, has chlorine as the limiting reactant.
The limiting reactant in a reaction has been given as the reactant that has been responsible for the amount of product formed.
The balanced chemical equation for the reaction has been:
[tex]\rm 2\;Al\;+\;3\;Cl_2\;\roghtarrow\;2\;AlCl_3[/tex]
The balanced chemical equation has been given that 2 moles of aluminum has been reacted for complete utilization of 3 moles of chlorine.
The moles of aluminum in 34 grams of sample has been:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}\\Moles\;Al=\dfrac{34}{26.98} \\Moles\;Al=1.26\;g/mol[/tex]
The available moles of Aluminum has been 1.26 g/mol.
The moles of Chlorine in 39 grams has been:
[tex]\rm Moles\;Cl=\dfrac{39}{35.45}\\Moles\;Cl=1.10\;g/mol[/tex]
The available moles of Chlorine has been 1.10 g/mol.
The moles of chlorine required to react with 1.26 g/mol Al has been:
[tex]\rm 2\;mole\;Al=3\;mol\;Cl\\1.26\;mol\;Al=\dfrac{3}{2}\;\times\;1.26\;mol\;Cl\\1.26\;mol\;Al=1.89\;mol\;Cl[/tex]
The required mole of Cl has been 1.89 mol. The available mol of Cl has been 1.10 mol. Thus, in the reaction of Cl and Al, Cl has been the limiting reactant.
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