Question

A 0.231 g piece of solid aluminum reacts with gaseous oxygen from the atmosphere to form solid aluminum oxide. In the laboratory, a student weighs the mass of the aluminum oxide collected from this reaction as 0.229 g. What is the percent yield of this reaction?

Answer 1

1. Let us first define the equation for the reaction :

4Al(s) + 3O2(g) → 2Al2O3 (s)

2. Now Let determine number of moles for ALand for Al2O3 :

• Number of moles for Aluminium (n) = mass of Al/Molecular Mass Al

= 0.231g ( given) / 26.982g/mol

= 8.56x10 ^-3 moles of Aluminium

• Number of moles for Al2O3 (n ) = mass of Al2O3/Molecular MassAL2O3

= (0.229 g. ) / 101.96g/mol

= 2.25x10^-3 moles of Al2O3

3. Calculate Percentage yield :

• According to the balanced equation above, 4 moles of AL : 2 moles AL2O3

Therefore :8.56x10 ^-3 moles of Al = 8.56x10 ^-3 mol * 2/4 = 4.28x10^-3 mol of Al2O3

• Theoritical yield : number of moles of AL2O3 * Molecular Mass of AL2O3

= 4.28x10^-3 *101.96g/mol

= 0.437 g

• Finally :

Percentage yield = actual yield /theoritical yield * 100

= 0.229 g/0.437 g *100

=68%