Question

When 3.8 g of aluminum are reacted with excess chlorine in the lab, you find you have made 17.8 g of aluminum chloride. What is the percent yield of this reaction?

Answer 1

Answer:

94.8% is the percentage yield of this reaction.

Explanation:

Mass of aluminium = 3.8 g

Moles of aluminium = $\frac{3.8 g}{27 g/mol}=0.1407 mol$

$2Al+3Cl_2\rightarrow 2AlCl_3$

According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride, then 0.1407 moles of aluminium will give:

$\frac{2}{2}\times 0.1407 mol=0.01407 mol$ 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:

$=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{17.8 g}{18.78 g}\times 100=94.8\%$

94.8% is the percentage yield of this reaction.