Question

A chemist burns 160.0 g of Al in O to produce Al2O3. She produces 260.0 g of Al2O3. Calculate the theoritical yield and percent yield

Answer 1

86.04 %

Step-by-step explanation:

Data given:

mass of Al = 160.0 g

actual yield Al₂O₃= 260 g

Theoretical yield = ?

percent yield ofAl₂O₃ = ?

Solution:

First we have to find theoretical yield.

So,

we Look at the balanced reaction

4Al + 3O₂ -----—> 2Al₂O₃

4 mol 2 mol

As 4 mole of Al give 2 mole of Al₂O₃

Convert moles to mass

molar mass of Al = 27 g/mol

molar mass of Al₂O₃ = 2(27) + 3(16)

molar mass of Al₂O₃ = 54 + 48

molar mass of Al₂O₃ = 102 g/mol

Now

4Al + 3O₂ -----—> 2Al₂O₃

4 mol (27g/mol) 2 mol (102 g/mol)

108 g 204 g

108 grams of Al produce 204 g of Al₂O₃

So

if 108 grams of Al produce 204 g of Al₂O₃ so how many grams of Al₂O₃ will be produced by 160 g of Al.

Apply Unity Formula

108 grams of Al ≅ 204 g of Al₂O₃

160 grams of Al ≅ X of Al₂O₃

Do cross multiply

mass of Al₂O₃= 204 g x 160 g / 108 g

mass of Al₂O₃ = 302.2 g

So the Theoretical yield of Al₂O₃ = 302.2 g

Now Find the percent yield of Al₂O₃

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 260g / 302.2 g x 100 %

percent yield = 86.04 %

percent yield of Al₂O₃ = 86.04 %