Respuesta :
Answer:
Theoretical yield = 13.7 g
% yield =76 %
Explanation:
For [tex]Al_2O_3[/tex]
Mass of [tex]Al_2O_3[/tex] = 4.07 g
Molar mass of [tex]Al_2O_3[/tex] = 101.96 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{4.07\ g}{101.96\ g/mol}[/tex]
[tex]Moles\ of\ Al_2O_3= 0.0399\ mol[/tex]
According to the reaction:
[tex]Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O[/tex]
1 mole of [tex]Al_2O_3[/tex] on reaction produces 1 mole of [tex]Al_2(SO_4)_3[/tex]
So,
0.0399 mole of [tex]Al_2O_3[/tex] on reaction produces 0.0399 mole of [tex]Al_2(SO_4)_3[/tex]
Moles of [tex]Al_2(SO_4)_3[/tex] obtained = 0.0399 mole
Molar mass of [tex]Al_2(SO_4)_3[/tex] = 342.2 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.0399= \frac{Mass}{342.2\ g/mol}[/tex]
[tex]Mass= 13.7\ g[/tex]
Theoretical yield = 13.7 g
The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-
[tex]\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100[/tex]
Given , Values from the question:-
Theoretical yield = 13.7 g
Experimental yield = 10.4 g
Applying the values in the above expression as:-
[tex]\%\ yield =\frac{10.4}{13.7}\times 100[/tex]
% yield =76 %
