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.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j