Achemist 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

86.04 %

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 %

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.