Respuesta :
Answer:
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%
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:
[tex]\rm Al(OH)_3\;+\;3\;HC\l\rightarrow\;AlCl_3\;+\;3\;H_2O[/tex]
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 = [tex]\rm \dfrac{Mass}{Molecular\;mass}[/tex]
Moles of Aluminum hydroxide = [tex]\rm \dfrac{14}{156}[/tex]
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 = [tex]\rm \dfrac{Mass}{266.68}[/tex]
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 = [tex]\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100[/tex]
The theoretical yield = 23.92 grams
The actual yield = 22 grams
% Yield = [tex]\rm \dfrac{22}{23.92}\;\times\;100[/tex]
% 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:
https://brainly.com/question/25199574
