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

94.8% is the percentage yield of this reaction.

Step-by-step explanation:

Mass of aluminium = 3.8 g

Moles of aluminium =
`(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:

`(2)/(2)* 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}}* 100`

`=(17.8 g)/(18.78 g)* 100=94.8\%`

94.8% is the percentage yield of this reaction.