Respuesta :
Answer:
The answer to your question is 1.49 mol of Cl₂
Explanation:
Data
mass of Al = 40.5 g
mass of Cl₂ = 212.7 g
moles of excess reactant = ?
- Balanced chemical reaction
2 Al(s) + 3Cl₂(g) ⇒ 2AlCl₃
Process
a) Calculate the molar mass of the reactants
molar mass of Al = 2 x 26.98 = 53.96 g
molar mass of Cl₂ = 6 x 35.45 = 212.7 g
b) Calculate the theoretical proportion Al/Cl₂ = 53.96/212.7 = 0.254
Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19
As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.
c) Calculate the grams of excess reactant
53.96 g of Al ------------------ 212.7 g of Cl₂
40.5 g of Al ------------------- x
x = (40.5 x 212.7) / 53.96
x = 8614.35 / 53.96
x = 159.64 g of Cl₂
Excess Cl₂ = 212.7 - 159.64
= 53.057 g
d) Calculate the moles of Cl
35.45 g of Cl ----------------- 1 mol
53.057 g of Cl --------------- x
x = (53.057 x 1)/35.45
x = 1.49 mol of Cl₂
Answer:
There will remain 0.75 moles of Cl2
Explanation:
Step 1: Data given
Mass of Al = 40.5 grams
Mass of Cl2 = 212.7 grams
Molar mass Al = 26.99 g/mol
Molar mass Cl2 = 70.9 g/mol
Step 2: The balanced equation
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
Step 3: Calculate moles
Moles = mass / molar mass
Moles Al = 40.5 grams / 26.99 g/mol
Moles Al = 1.50 moles
Moles Cl2 = 212.7 grams / 70.9 g/mol
Moles Cl2 = 3.0 moles
Step 4: Calculate limiting reactant
For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3
Al is the limiting reactant. It will completely be consumed (1.50 moles)
Cl2 is in excess. There will react 3/2 * 1.50 = 2.25 moles
There will remain 3.0 - 2.25 = 0.75 moles
There will remain 0.75 moles of Cl2
