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If you react 125 g of AlCl₃ with 65.0 g of NaOH in solution, what is the theoretical yield, in grams, of aluminum hydroxide?

If only 35.8 g of aluminum hydroxide are recovered, what is the percent yield?

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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