Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.