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.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"