Question

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: In a particular experiment, the reaction of 2.5 g of Al with 2.5 g of O2 produced 3.5 g of Al2O3. The % yield of the reaction is __________.

Answer 1

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 = $\frac{actual yield }{theoretical yield}$ ×100 = $\frac{3.5}{4.74}$ × 100 = 73.8 % yield