Answer :
The percent yield of aluminum hydroxide is approximately 49.1%.
The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the moles of each reactant using their molar masses:
Molar mass of AlCl3 = 133.34 g/mol
Molar mass of NaOH = 39.997 g/mol
Moles of AlCl3 = 125 g / 133.34 g/mol
Moles of NaOH = 65.0 g / 39.997 g/mol
Now, determine the stoichiometric ratio between AlCl3 and Al(OH)3 from the balanced equation. assume it is a 1:1 ratio:
Moles of Al(OH)3 formed = Moles of AlCl3 (limiting reactant)
Next, calculate the theoretical yield of aluminum hydroxide (Al(OH)3) using its molar mass:
Theoretical yield of Al(OH)3 = Moles of Al(OH)3 formed * Molar mass of Al(OH)3
Now, calculate the percent yield using the experimental yield and theoretical yield:
Percent yield = (Experimental yield / Theoretical yield) * 100%
Given:
Experimental yield = 35.8 g
Performing the calculations, get:
Moles of AlCl3 = 125 g / 133.34 g/mol ≈ 0.937 mol
Moles of NaOH = 65.0 g / 39.997 g/mol ≈ 1.625 mol
Assuming a 1:1 stoichiometric ratio, the moles of Al(OH)3 formed would be 0.937 mol.
Theoretical yield of Al(OH)3 = 0.937 mol * (78.0 g/mol) ≈ 73.026 g
Percent yield = (35.8 g / 73.026 g) * 100% ≈ 49.1%
Therefore, the percent yield of aluminum hydroxide is approximately 49.1%.
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