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

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.

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 =

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

=

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 = x 100

                      = x 100

  percent yield = 92.8 % is the yield percent.

the answer is barium and copper bromide

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2