Respuesta :
Answer:
The % yield of the reaction is 73.8 %
Explanation:
To solve this, we list out the given variables thus
Mass of aluminium in the experiment = 2.5 g
mass of oxygen gas in the experiment = 2.5 g
Molar mass of aluminium = 26.98 g/mol
molar mass of oxygen O₂ = 32 g/mol
The reaction between aluminium and gaseous oxygen can be written as follows
4Al + 3O₂ → 2Al₂O₃
Thus four moles of aluminium forms two moles of aluminium oxide
Thus (2.5 g)÷(26.98 g/mol) = 0.093 mole of aluminium and
(2.5 g)÷(32 g/mol) = 0.078125 moles of oxygen
However four moles of aluminium react with three moles of oxygen gas O₂
1 mole of aluminum will react with 3/4 moles of oxygen O₂ and 0.093 mole of aluminium will react with 0.093*3/4 moles of O₂ = 0.0695 moles of Oxygen hence aluminium is the limiting reagent and we have
1 mole of oxygen will react with 4/3 mole of aluminium
∴ 0.078125 mole of oxygen will react with 0.104 moles of aluminium
Therefore 0.093 mole of aluminium will react with O₂ to produce 2/4×0.093 or 0.0465 moles of 2Al₂O₃
The molar mass of 2Al₂O₃ = 101.96 g/mol
Hence the mass of 0.0465 moles = number of moles × (molar mass)
= 0.0465 moles × 101.96 g/mol = 4.74 g
The of aluminium oxide Al₂O₃ is 4.74 g, but the actual yield = 3.5 g
Therefore the Percentage yield = [tex]\frac{actual yield }{theoretical yield}[/tex] ×100 = [tex]\frac{3.5}{4.74}[/tex] × 100 = 73.8 % yield
