Question

If 2.87 g of aluminum are reacted with excess copper (II) sulfate and 9.2 g of copper are produced what is the percent yield

Answer 1

92.8 % is the percent yield when 2.87 g of aluminum are reacted with excess copper (II) sulfate and 9.2 g of copper are produced.

Step-by-step explanation:

Mass of aluminum = 2.87 grams

mass of copper produced = 9.2 grams ( actual yield)

the balanced chemical reaction is given as:

2 Al + 3 CuSO4 ⇒ ( AL2(SO4)3 +3 Cu.

from the reaction it is found that

2 moles of aluminum reacted to give 3 moles of copper

so converting them in to mass.

number of moles of copper in 2.8 grams =
`(2.8)/(63.4)`

number of moles = 0.104 moles of copper

atomic mass of aluminum = 26.98 grams/mole

atomic mass of copper = 63.54 grams/mole

From the given data

2.87 grams Al gives 9.2 grams of Cu.

2 moles of aluminum reacted to give 3 moles of copper

0.104 moles of aluminum will give

`(3)/(2)` =
`(x)/(0.104)`

2x = 0.132

x = 0.156 moles of Cu will be formed.

Hence theoretical yield should be = 0.156 x 63.54

= 9.91 grams

Percent yield =
`(actual yield)/(theoretical yield)` x 100

=
`(9.2)/(9.91)` x 100

percent yield = 92.8 % is the yield percent.