Answer :
Let's proceed step by step to solve this problem and fill in the requested BCA (Before, Change, After) table.
### Step 1: Calculate the Moles of Reactants
We start with the given masses and molar masses:
- Mass of Al = 4.20 g
- Molar mass of Al = 26.98 g/mol
- Mass of I₂ = 17.40 g
- Molar mass of I₂ = 253.80 g/mol
First, determine the moles of Al:
[tex]$ \text{Moles of Al} = \frac{4.20 \, \text{g}}{26.98 \, \text{g/mol}} = 0.15567086730911786 \, \text{mol} $[/tex]
Next, determine the moles of I₂:
[tex]$ \text{Moles of I₂} = \frac{17.40 \, \text{g}}{253.80 \, \text{g/mol}} = 0.0685579196217494 \, \text{mol} $[/tex]
### Step 2: Determine the Limiting Reagent
From the balanced chemical equation:
[tex]$ 2 \text{Al} + 3 \text{I}_2 \rightarrow 2 \text{AlI}_3 $[/tex]
The stoichiometric ratio is 2 moles of Al to 3 moles of I₂.
Calculate the required amount of I₂ for the given moles of Al:
[tex]$ \text{Required moles of I₂} = \frac{3}{2} \times 0.15567086730911786 \, \text{mol} = 0.23350630096367679 \, \text{mol} $[/tex]
Since 0.2335 mol of I₂ is required but only 0.0686 mol is available, I₂ is the limiting reagent.
### Step 3: Use the Limiting Reagent to Determine the Maximum Product
Since I₂ is the limiting reagent, we determine the moles of product based on I₂.
For the balanced reaction, 3 moles of I₂ produce 2 moles of AlI₃. Therefore:
[tex]$ \text{Moles of AlI}_3 = \frac{2}{3} \times 0.0685579196217494 \, \text{mol} = 0.04570527974783293 \, \text{mol} $[/tex]
### Step 4: Calculate the Mass of AlI₃ Produced
The molar mass of AlI₃ is:
[tex]$ \text{Molar mass of AlI}_3 = 2 \times 26.98 \, \text{g/mol} + 3 \times 126.90 \, \text{g/mol} = 407.58 \, \text{g/mol} $[/tex]
Thus, the mass of AlI₃ produced is:
[tex]$ \text{Mass of AlI}_3 = 0.04570527974783293 \, \text{mol} \times 407.58 \, \text{g/mol} = 18.636 \, \text{g} $[/tex]
### Step 5: Fill in the BCA Table
Let’s fill the table assuming the limiting reagent is I₂:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \ & 2 \text{Al} & + & 3 \text{I}_2 & \rightarrow & 2 \text{AlI}_3 \\ \hline \text{Before (mol)} & 0.1557 & & 0.0686 & & 0 \\ \hline \text{Change (mol)} & -0.1038 & & -0.0686 & & +0.0457 \\ \hline \text{After (mol)} & 0.0519 & & 0 & & 0.0457 \\ \hline \end{array} \][/tex]
Here, the changes are calculated based on the stoichiometry of the reaction:
- For every 3 moles of I₂, 2 moles of AlI₃ are produced and 2 moles of Al are consumed.
- Since the initial moles of I₂ are 0.0686, all of it reacts and 0.0686 mol of I₂ produces 0.0457 mol of AlI₃ and consumes 0.1038 mol of Al.
All this confirms that the total mass of the product (AlI₃) produced is 18.636 g.
### Step 1: Calculate the Moles of Reactants
We start with the given masses and molar masses:
- Mass of Al = 4.20 g
- Molar mass of Al = 26.98 g/mol
- Mass of I₂ = 17.40 g
- Molar mass of I₂ = 253.80 g/mol
First, determine the moles of Al:
[tex]$ \text{Moles of Al} = \frac{4.20 \, \text{g}}{26.98 \, \text{g/mol}} = 0.15567086730911786 \, \text{mol} $[/tex]
Next, determine the moles of I₂:
[tex]$ \text{Moles of I₂} = \frac{17.40 \, \text{g}}{253.80 \, \text{g/mol}} = 0.0685579196217494 \, \text{mol} $[/tex]
### Step 2: Determine the Limiting Reagent
From the balanced chemical equation:
[tex]$ 2 \text{Al} + 3 \text{I}_2 \rightarrow 2 \text{AlI}_3 $[/tex]
The stoichiometric ratio is 2 moles of Al to 3 moles of I₂.
Calculate the required amount of I₂ for the given moles of Al:
[tex]$ \text{Required moles of I₂} = \frac{3}{2} \times 0.15567086730911786 \, \text{mol} = 0.23350630096367679 \, \text{mol} $[/tex]
Since 0.2335 mol of I₂ is required but only 0.0686 mol is available, I₂ is the limiting reagent.
### Step 3: Use the Limiting Reagent to Determine the Maximum Product
Since I₂ is the limiting reagent, we determine the moles of product based on I₂.
For the balanced reaction, 3 moles of I₂ produce 2 moles of AlI₃. Therefore:
[tex]$ \text{Moles of AlI}_3 = \frac{2}{3} \times 0.0685579196217494 \, \text{mol} = 0.04570527974783293 \, \text{mol} $[/tex]
### Step 4: Calculate the Mass of AlI₃ Produced
The molar mass of AlI₃ is:
[tex]$ \text{Molar mass of AlI}_3 = 2 \times 26.98 \, \text{g/mol} + 3 \times 126.90 \, \text{g/mol} = 407.58 \, \text{g/mol} $[/tex]
Thus, the mass of AlI₃ produced is:
[tex]$ \text{Mass of AlI}_3 = 0.04570527974783293 \, \text{mol} \times 407.58 \, \text{g/mol} = 18.636 \, \text{g} $[/tex]
### Step 5: Fill in the BCA Table
Let’s fill the table assuming the limiting reagent is I₂:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \ & 2 \text{Al} & + & 3 \text{I}_2 & \rightarrow & 2 \text{AlI}_3 \\ \hline \text{Before (mol)} & 0.1557 & & 0.0686 & & 0 \\ \hline \text{Change (mol)} & -0.1038 & & -0.0686 & & +0.0457 \\ \hline \text{After (mol)} & 0.0519 & & 0 & & 0.0457 \\ \hline \end{array} \][/tex]
Here, the changes are calculated based on the stoichiometry of the reaction:
- For every 3 moles of I₂, 2 moles of AlI₃ are produced and 2 moles of Al are consumed.
- Since the initial moles of I₂ are 0.0686, all of it reacts and 0.0686 mol of I₂ produces 0.0457 mol of AlI₃ and consumes 0.1038 mol of Al.
All this confirms that the total mass of the product (AlI₃) produced is 18.636 g.
