Aluminum hydroxide is often present in antacids to neutralize stomach acid (HCl). If 14.0 g aluminum hydroxide is present in an antacid tablet, determine the theoretical yield of aluminum chloride produced when the tablet reacts with stomach acid. If the actual yield of the aluminum chloride from this tablet is 22.0 g, what is the percent yield?

The percent yield of aluminum chloride in the neutralization reaction has been 91.97%.

The balanced reaction for neutralization of stomach acid with aluminum hydroxide has been:

The reaction has been determined that 1 mole of aluminum hydroxide forms 1 mole of aluminum chloride. The moles of aluminum hydroxide has been given as:

Moles =  

Moles of Aluminum hydroxide =

Moles of Aluminum hydroxide = 0.0897 mol.

Thus, the aluminum chloride formed has been:

1 mole Aluminum hydroxide = 1 mole aluminum chloride

0.0897 mole aluminum hydroxide = 0.0897 mole aluminum chloride

The mass of Aluminum chloride formed has been:

0.0897 mole aluminum chloride =

Mass of Aluminum chloride = 0.0897 mol × 266.68 g/mol

Mass of Aluminum chloride = 23.92 grams

The %yield can be given as:

%Yield =

The theoretical yield = 23.92 grams

The actual yield = 22 grams

% Yield =

% Yield = 91.97%.

The percent yield of aluminum chloride in the neutralization reaction has been 91.97%.

For more information about the neutralization reaction, refer to the link:

link

Percent Yield = 91.97%

Explanation:

Aluminum hydroxide Formula: Al(OH)3 Molar mass: 78 g/mol

Stomach acid Formula: HCl Molar mass: 36.46 g/mol

Aluminum Chloride Formula: AlCl₃ Molar mass: 133.34 g/mol

14g Al(OH)3

2Al(OH)3 + Cl2 → 2Al(OH)3Cl

Balanced equation

14/156 = 0.0897 Moles of Al(OH)3

Ratio is 2:1:2 so same ratio for Al(OH)3 to AlCl₃

0.0897 x 266.68 = 23.92grams of AlCl₃

Theoretical yield of aluminum chloride

Actual yield aluminum chloride 22.0 g

Percent yield?

Percent Yield = (actual Yield/Theoretical yield) x 100

Percent Yield = (22/23.92) x 100

Percent Yield = 91.97%

wher is the rest of the question