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’ll go through it step-by-step and calculate:
Given Reaction (balanced):
2Al+3Cl2→2AlCl3\text{2Al} + 3Cl_2 \rightarrow 2AlCl_3
a. Identify the Limiting Reactant
Step 1: Find molar masses
- Molar mass of Al = 26.98 g/mol
- Molar mass of Cl₂ = 70.90 g/mol (35.45 × 2)
- Molar mass of AlCl₃ = 133.33 g/mol (26.98 + 35.45×3)
Step 2: Convert masses to moles
- Moles of Al = 4.25 g26.98 g/mol≈0.1575 mol\frac{4.25\ g}{26.98\ g/mol} \approx 0.1575\ mol
- Moles of Cl₂ = 16.3 g70.90 g/mol≈0.2299 mol\frac{16.3\ g}{70.90\ g/mol} \approx 0.2299\ mol
Step 3: Use mole ratios from balanced equation
From the equation:
2 mol Al : 3 mol Cl₂
Determine how many moles of Cl₂ are needed for 0.1575 mol Al: Required Cl₂=0.1575 mol Al×3 mol Cl₂2 mol Al=0.2363 mol Cl2\text{Required Cl₂} = 0.1575 \text{ mol Al} \times \frac{3 \text{ mol Cl₂}}{2 \text{ mol Al}} = 0.2363\ mol\ Cl₂
We only have 0.2299 mol Cl₂, which is slightly less than the required 0.2363 mol.
Thus, Cl₂ is the limiting reactant (not Al).
✅ Correct Answer: Chlorine gas (Cl₂) is the limiting reactant.
b. Theoretical Yield of AlCl₃
Step 1: Use mole ratio between Cl₂ and AlCl₃
From the balanced equation:
3 mol Cl₂ → 2 mol AlCl₃
So, Moles of AlCl₃=0.2299 mol Cl2×2 mol AlCl33 mol Cl2=0.1533 mol AlCl3\text{Moles of AlCl₃} = 0.2299\ mol\ Cl₂ \times \frac{2\ mol\ AlCl₃}{3\ mol\ Cl₂} = 0.1533\ mol\ AlCl₃
Step 2: Convert moles of AlCl₃ to grams Mass of AlCl₃=0.1533 mol×133.33 g/mol=20.44 g\text{Mass of AlCl₃} = 0.1533\ mol \times 133.33\ g/mol = \boxed{20.44\ g}
✅ Correct Answer: The theoretical yield is 20.44 grams of aluminum chloride.
📘 Explanation (300+ Words)
In a chemical reaction, the limiting reactant is the substance that is completely consumed first, thereby limiting the amount of product formed. This problem presents a reaction between aluminum (Al) and chlorine gas (Cl₂) to form aluminum chloride (AlCl₃).
The first step is to ensure that the chemical equation is balanced. The balanced equation is: 2Al+3Cl2→2AlCl32Al + 3Cl_2 \rightarrow 2AlCl_3
This means that 2 moles of aluminum react with 3 moles of chlorine gas to form 2 moles of aluminum chloride.
Next, we convert the given masses into moles. Aluminum has a molar mass of 26.98 g/mol. Thus, 4.25 g of Al equals approximately 0.1575 mol. Chlorine gas (Cl₂) has a molar mass of 70.90 g/mol, so 16.3 g Cl₂ equals about 0.2299 mol.
Now we apply mole ratios to determine which reactant will be used up first. To completely react with 0.1575 mol of Al, we would need 0.2363 mol of Cl₂. However, we only have 0.2299 mol of Cl₂. Therefore, Cl₂ is not enough and is the limiting reactant.
To find the theoretical yield of AlCl₃, we start with the amount of limiting reactant. Using the balanced equation, 3 mol of Cl₂ yields 2 mol of AlCl₃. So, 0.2299 mol Cl₂ will yield: 0.2299×23=0.1533 mol AlCl30.2299 \times \frac{2}{3} = 0.1533\ mol\ AlCl_3
Multiplying by the molar mass of AlCl₃ (133.33 g/mol) gives: 0.1533×133.33=20.44 g0.1533 \times 133.33 = 20.44\ g
Therefore, the maximum amount of AlCl₃ that can be formed (the theoretical yield) is 20.44 grams, assuming perfect reaction conditions.