Balance (6) 13. If 4.25 g of aluminum reacts with 16.3 g of chlorine gas to produce aluminum chloride according to the following unbalanced equation, 2 Al + 3Cl2 2 AICI: a. Which chemical is the limiting reactant? (Show your calculations) Al-26.98×2 = 53.96 = 1.63% a-35.45 x6 = 212.3 = 6.11% 266.60 Al is the limiting rencant because it 4.25 will un act ofter 4 asg. Its the chemical 1621 with the smallest amung b. What is the theoretical yield (in grams) of aluminum chloride?
The Correct Answer and Explanation is:
To solve this problem, we must go through several steps: balancing the chemical equation, determining the limiting reactant, and then calculating the theoretical yield of aluminum chloride (AlCl₃).
Balanced Equation
2Al+3Cl2→2AlCl32Al + 3Cl_2 \rightarrow 2AlCl_32Al+3Cl2→2AlCl3
Step a: Determine the Limiting Reactant
We are given:
- 4.25 g of Aluminum (Al)
- 16.3 g of Chlorine gas (Cl₂)
1. Molar Masses:
- Molar mass of Al = 26.98 g/mol
- Molar mass of Cl₂ = 70.90 g/mol (because Cl = 35.45, and Cl₂ = 2 × 35.45)
2. Convert mass to moles:
- Moles of Al:
4.25 g26.98 g/mol=0.1575 mol\frac{4.25 \, \text{g}}{26.98 \, \text{g/mol}} = 0.1575 \, \text{mol}26.98g/mol4.25g=0.1575mol
- Moles of Cl₂:
16.3 g70.90 g/mol=0.2299 mol\frac{16.3 \, \text{g}}{70.90 \, \text{g/mol}} = 0.2299 \, \text{mol}70.90g/mol16.3g=0.2299mol
3. Determine limiting reactant using stoichiometry:
From the balanced equation: 2 mol Al:3 mol Cl2⇒0.15752=0.07875and0.22993=0.076632 \text{ mol Al} : 3 \text{ mol Cl}_2 \Rightarrow \frac{0.1575}{2} = 0.07875 \quad \text{and} \quad \frac{0.2299}{3} = 0.076632 mol Al:3 mol Cl2⇒20.1575=0.07875and30.2299=0.07663
Since 0.07663 < 0.07875, chlorine gas (Cl₂) limits the reaction.
Step b: Theoretical Yield of AlCl₃
From the balanced equation: 3 mol Cl2→2 mol AlCl33 \text{ mol Cl}_2 \rightarrow 2 \text{ mol AlCl}_33 mol Cl2→2 mol AlCl3
Using limiting reactant (Cl₂):
- Moles of AlCl₃ produced:
0.2299 mol Cl2×2 mol AlCl33 mol Cl2=0.1533 mol AlCl30.2299 \, \text{mol Cl}_2 \times \frac{2 \, \text{mol AlCl}_3}{3 \, \text{mol Cl}_2} = 0.1533 \, \text{mol AlCl}_30.2299mol Cl2×3mol Cl22mol AlCl3=0.1533mol AlCl3
Molar mass of AlCl₃:
- Al = 26.98
- Cl = 35.45 × 3 = 106.35
- Total = 133.33 g/mol
Mass of AlCl₃ produced: 0.1533 mol×133.33 g/mol=20.44 g0.1533 \, \text{mol} \times 133.33 \, \text{g/mol} = \boxed{20.44 \, \text{g}}0.1533mol×133.33g/mol=20.44g
Final Answers:
- Limiting Reactant: Chlorine gas (Cl₂)
- Theoretical Yield of AlCl₃: 20.44 g
Explanation (300 words):
To find the theoretical yield in a chemical reaction, it’s essential first to identify the limiting reactant, which is the substance that runs out first and thus limits the amount of product that can form.
In this reaction, aluminum reacts with chlorine gas to form aluminum chloride according to the balanced equation: 2Al+3Cl2→2AlCl32Al + 3Cl_2 \rightarrow 2AlCl_32Al+3Cl2→2AlCl3
We start by converting the given masses of aluminum (4.25 g) and chlorine gas (16.3 g) into moles using their molar masses. Aluminum has a molar mass of 26.98 g/mol, and chlorine gas (Cl₂) has a molar mass of 70.90 g/mol. This gives us 0.1575 mol of Al and 0.2299 mol of Cl₂.
Using the stoichiometric ratio from the balanced equation (2 mol Al : 3 mol Cl₂), we compare how much of each reactant would be needed. When we divide the moles of each reactant by its stoichiometric coefficient, we find that chlorine gas gives the smaller result. This means chlorine is the limiting reactant, because it will be consumed before all the aluminum reacts.
Next, we use the moles of the limiting reactant (Cl₂) to determine how much product (AlCl₃) can form. The molar ratio tells us that 3 mol of Cl₂ produces 2 mol of AlCl₃. Applying this ratio to 0.2299 mol of Cl₂, we calculate that 0.1533 mol of AlCl₃ can form. Finally, converting moles of AlCl₃ into grams using its molar mass (133.33 g/mol), we find the theoretical yield is 20.44 grams.