Answer:
94.8% is the percentage yield of this reaction.
Explanation:
Mass of aluminium = 3.8 g
Moles of aluminium = [tex]\frac{3.8 g}{27 g/mol}=0.1407 mol[/tex]
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride, then 0.1407 moles of aluminium will give:
[tex]\frac{2}{2}\times 0.1407 mol=0.01407 mol[/tex] aluminium chloride
Mass of 0.1407 moles of aluminum chloride:
= 0.1407 mol × 133.5 g/mol = 18.78 g
Theoretical yield of aluminum chloride = 18.78 g
Experimental yield of aluminum chloride = 17.8 g
The percentage yield of reaction:
[tex]=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{17.8 g}{18.78 g}\times 100=94.8\%[/tex]
94.8% is the percentage yield of this reaction.
