17 kg of aluminium was produced from 51 kg of aluminium oxide (al2o3) by electrolysis. what was the percentage yield? (ar : al 5 27, o 5 16.)
Answers
The overall equation for the electrolysis of aluminium oxide (AlO3) is as
follows:
aluminium oxide aluminium + oxygen
2Al₂O₃ (l) ⇒ 4Al (l)
+ 3O₂ (g)
Using stochimetry, we
can calculate the theoretical yield of aluminium from the electrolysis of
aluminium oxide.
Here are the steps:
The mole ratio in the equation between Al₂O₃ :
Al is 2 : 4
Calculate the moles of 51kg of Al₂O₃
Moles = mass/molar mass
mass = 51kg (5100g)
molar mass = 102
moles = 5100/102 = 50 moles
Remember the mole
ratio between Al₂O₃:Al is 2 : 4
Therefore theoretical moles of Al produced from 51kg of Al₂O₃ is:
4/2 × 50= 100 moles
Therefore
theoretical mole yield of Al is 100 moles
Find mass of 1 mole of Al
Mass = moles × molar mass
moles = 100
molar mass = 27
Therefore the mass = 27 × 100 = 2700g
Calculate the percentage yield:
% yield = actual yield/theoretical yield ×100%
actual yield = 17kg
Theoretical yield = 27kg
17kg/27kg × 100% = 0.6296
= 62.96%
Hence the % yield = 62.96%
Answer:
Explanation:
santy's answer is correct, but 51000/102 is 500 moles, not 50 moles