Question

8. A chemist burns 160.0 g of Al in 234.78 g of oxygen to produce aluminum oxide. She produces 260.0g of solid aluminum oxide. Write a balanced equation for this reaction. Calculate the percent yield. (3 pts)

Answer 1

1)

Balanced chemical equation is:

4 Al + 2 O2 ---> 2 Al2O3

2)

Molar mass of Al = 26.98 g/mol

mass(Al)= 160.0 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(1.6*10^2 g)/(26.98 g/mol)

= 5.93 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 234.78 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(2.348*10^2 g)/(32 g/mol)

= 7.337 mol

  

4 mol of Al reacts with 2 mol of O2

for 5.93 mol of Al, 2.965 mol of O2 is required

But we have 7.337 mol of O2

so, Al is limiting reagent

we will use Al in further calculation

Molar mass of Al2O3,

MM = 2*MM(Al) + 3*MM(O)

= 2*26.98 + 3*16.0

= 101.96 g/mol

According to balanced equation

mol of Al2O3 formed = (2/4)* moles of Al

= (2/4)*5.93

= 2.965 mol

use:

mass of Al2O3 = number of mol * molar mass

= 2.965*1.02*10^2

= 3.023*10^2 g

% yield = actual mass*100/theoretical mass

= 2.6*10^2*100/3.023*10^2

= 86.0 %

Answer: 86.0 %