Respuesta :
Answer:
Mass = 5.33 g
Explanation:
Given data:
Mass of Al = 2.80 g
Mass of Cl₂ = 4.15 g
Theoretical yield of AlCl₃ = ?
Solution:
Chemical equation:
2Al + 3Cl₂ → 2AlCl₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 2.80 g/ 27 g/mol
Number of moles = 0.10 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.15 g/71 g/mol
Number of moles = 0.06 mol
Now we will compare the moles of AlCl₃ with Al and Cl₂.
Cl₂ : AlCl₃
3 : 2
0.06 : 2/3×0.06 = 0.04
Al : AlCl₃
2 : 2
0.10 : 0.10
Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.
Mass of AlCl₃:Theoretical yield
Mass = number of moles ×molar mass
Mass = 0.04 mol × 133.34 g/mol
Mass = 5.33 g
5.33 grams is the theoretical yield of AlCl₃ in this reaction.
What is theoretical yield?
Theoretical yield of any reaction tells about the accurate quantity of the product formed by the complete consumption of reactants.
Given chemical reaction is:
2Al + 3Cl₂ → 2AlCl₃
First we convert the mass of reactants into moles by using the formulas:
n = W/M, where
W = given mass
M = molar mass
Given mass of Al = 2.80 g
Moles of Al = 2.80/27 = 0.10 mole
Given mass of Cl₂ = 4.15 g
Moles of Cl₂ = 4.15/71 = 0.06 mole
From this calculation we conclude that Cl₂ is the limiting reagent, so the formation of product depends on Cl₂.
From the stoichiometry of the reaction, it is clear that:
3 moles of Cl₂ = produce 2 moles of AlCl₃
0.06 moles of Cl₂ = produce 2/3×0.06 = 0.04 moles of AlCl₃
Now we calculate the produced mass of AlCl₃ as:
W = 0.04mol × 133.34g/mol = 5.33g
Hence, 5.33g is the produced theoretical yield.
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