Question

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

Answer 1

I have not a clue sorry...  :(

Answer 2

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

`Mole = (given mass (g))/(atomic mass (g)/(mol))`

`Mole = (35.5 g )/(26.9 (g)/(mol))`

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

`Mole of AlCl_3 = Mole of Al * Mole ratio`

`Mole of AlCl_3 = 1.32 mol of Al * (2)/(2)`

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ =
`133.34 (g)/(mol)`

Mass of AlCl3 can be calculated using mole formula as:

`1.32 mol of AlCl_3 = ( mass (g))/(133.34 (g)/(mol))`

Multiplying both side by
`133.34 (g)/(mol)`

`1.32 mole * 133.34(g)/(mol) = (mass (g))/(133.34(g)/(mol)) *133.34 (g)/(mol)`

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

`Mole of Cl_2 = (39.0 g)/(70.9(g)/(mol))`

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

`Mole of AlCl_3 = mole of Cl_2 * mole ratio`

`Mole of AlCl_3 = 0.55 mole * (2)/(3)`

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

`0.367 mol of AlCl_3 = (mass (g) )/(133.34 (g)/(mol))`

Multiplying both side by

`133.34 (g)/(mol)`

`0.367 mol of AlCl_3 * 133.34 (g)/(mol) = (mass(g))/(133.34(g)/(mol)) * 133.34 (g)/(mol)`

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .