Question

question 28
aluminum metal (1.63 g) reacts with 1.60 l of chlorine gas at 21 °c and 863 mmhg to produce 4.52 g of solid aluminum chloride.
start by writing a balanced chemical equation for this process, complete with phase symbols:
→
calculate the percent yield for this reaction.
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Explanation:

Step1: Write Balanced Equation

Aluminum (Al, solid) reacts with chlorine gas ($\ce{Cl2}$, gas) to form aluminum chloride ($\ce{AlCl3}$, solid). The balanced equation is:
$$2\ce{Al}(s) + 3\ce{Cl2}(g) ightarrow 2\ce{AlCl3}(s)$$

Step2: Find Moles of Al

Molar mass of Al = 26.98 g/mol.
Moles of Al = $\frac{1.63\ \text{g}}{26.98\ \text{g/mol}} \approx 0.0604\ \text{mol}$.

Step3: Find Moles of $\ce{Cl2}$ (Ideal Gas Law)

Ideal gas law: $PV = nRT$.
Convert $T$ to K: $21^\circ\text{C} + 273.15 = 294.15\ \text{K}$.
Convert $P$ to atm: $863\ \text{mmHg} \times \frac{1\ \text{atm}}{760\ \text{mmHg}} \approx 1.135\ \text{atm}$.
$V = 1.60\ \text{L}$, $R = 0.0821\ \text{L·atm/(mol·K)}$.
$n = \frac{PV}{RT} = \frac{1.135\ \text{atm} \times 1.60\ \text{L}}{0.0821\ \text{L·atm/(mol·K)} \times 294.15\ \text{K}} \approx 0.0748\ \text{mol}$.

Step4: Determine Limiting Reactant

From balanced equation, 2 mol Al reacts with 3 mol $\ce{Cl2}$.
Moles of $\ce{Cl2}$ needed for Al: $0.0604\ \text{mol Al} \times \frac{3\ \text{mol Cl2}}{2\ \text{mol Al}} = 0.0906\ \text{mol Cl2}$.
We have only 0.0748 mol $\ce{Cl2}$, so $\ce{Cl2}$ is limiting.

Step5: Calculate Theoretical Yield of $\ce{AlCl3}$

From balanced equation, 3 mol $\ce{Cl2}$ produces 2 mol $\ce{AlCl3}$.
Moles of $\ce{AlCl3}$ theoretical = $0.0748\ \text{mol Cl2} \times \frac{2\ \text{mol AlCl3}}{3\ \text{mol Cl2}} \approx 0.0499\ \text{mol}$.
Molar mass of $\ce{AlCl3}$ = 26.98 + 3×35.45 = 133.33 g/mol.
Theoretical yield = $0.0499\ \text{mol} \times 133.33\ \text{g/mol} \approx 6.65\ \text{g}$? Wait, no—wait, recalculate:
Wait, 0.0748 mol $\ce{Cl2}$ × (2 mol $\ce{AlCl3}$ / 3 mol $\ce{Cl2}$) = 0.04987 mol $\ce{AlCl3}$.
0.04987 mol × 133.33 g/mol ≈ 6.65 g? But wait, maybe I mixed up limiting reactant. Wait, let's check Al: 0.0604 mol Al × (2 mol $\ce{AlCl3}$ / 2 mol Al) = 0.0604 mol $\ce{AlCl3}$. But $\ce{Cl2}$ is limiting, so use $\ce{Cl2}$. Wait, no: 3 mol $\ce{Cl2}$ → 2 mol $\ce{AlCl3}$, so 1 mol $\ce{Cl2}$ → 2/3 mol $\ce{AlCl3}$. So 0.0748 mol $\ce{Cl2}$ → 0.0748 × (2/3) ≈ 0.0499 mol $\ce{AlCl3}$. Molar mass of $\ce{AlCl3}$: Al=26.98, Cl=35.45×3=106.35, total=26.98+106.35=133.33 g/mol. So theoretical yield = 0.0499 mol × 133.33 g/mol ≈ 6.65 g? Wait, but the actual yield is 4.52 g. Wait, maybe I made a mistake in limiting reactant. Let's recheck:

Moles of Al: 1.63 / 26.98 ≈ 0.0604 mol.
Moles of $\ce{Cl2}$: 0.0748 mol.

Ratio of Al to $\ce{Cl2}$ in equation: 2:3. So actual ratio: 0.0604 : 0.0748 ≈ 0.807 : 1.
Stoichiometric ratio: 2/3 ≈ 0.666 : 1.
Since actual ratio (Al:$\ce{Cl2}$) is greater than stoichiometric (0.807 > 0.666), Al is in excess, $\ce{Cl2}$ is limiting. Correct.

So theoretical yield from $\ce{Cl2}$: 0.0748 mol $\ce{Cl2}$ × (2 mol $\ce{AlCl3}$ / 3 mol $\ce{Cl2}$) = 0.04987 mol $\ce{AlCl3}$.
Theoretical mass: 0.04987 mol × 133.33 g/mol ≈ 6.65 g. Wait, but let's check with Al: 0.0604 mol Al × (2 mol $\ce{AlCl3}$ / 2 mol Al) = 0.0604 mol $\ce{AlCl3}$. But since $\ce{Cl2}$ is limiting, we use $\ce{Cl2}$'s moles.

Wait, but maybe I miscalculated moles of $\ce{Cl2}$. Let's recalculate $n$:

$P = 863\ \text{mmHg} = 863 / 760 ≈ 1.1355\ \text{atm}$.
$V = 1.60\ \text{L}$, $T = 294.15\ \text{K}$, $R = 0.0821$.

$n = (1.1355 × 1.60) / (0.0821 × 294.15) = (1.8168) / (24.15) ≈ 0.0752\ \text{mol}$. Ah, earlier was 0.0748, close enough.

So theoretical yield: 0.0752 mol $\ce{Cl2}$ × (2/3) = 0.0501 mol $\ce{AlCl3}$.
0.0501 mol × 133.33 g/mol ≈ 6.68 g.

Actual yield is 4.52 g.

Percent yield = (actual / theoretical) × 100 = (4.52 / 6.68) × 100 ≈ 67.7%? Wait, no—wait, maybe I messed up the limiting reactant. Wait, let's check Al's moles: 1.63 g Al is 0.0604 mol. The equation is 2 Al + 3 Cl2 → 2 AlCl3. So 2 mol Al needs 3 mol Cl2. So 0.0604 mol Al needs (3/2)×0.0604 = 0.0906 mol Cl2. But we have only 0.0752 mol Cl2, which is less than 0.0906, so Cl2 is limiting. Correct. So theoretical yield is based on Cl2.

Wait, but let's calculate theoretical yield from Cl2: 3 mol Cl2 → 2 mol AlCl3. So 1 mol Cl2 → 2/3 mol AlCl3. So 0.0752 mol Cl2 → 0.0752 × (2/3) = 0.0501 mol AlCl3. Mass: 0.0501 mol × 133.33 g/mol = 6.68 g.

Actual yield: 4.52 g.

Percent yield = (4.52 / 6.68) × 100 ≈ 67.7%? Wait, but maybe I made a mistake in molar mass. AlCl3: Al=26.98, Cl=35.45×3=106.35, total=26.98+106.35=133.33 g/mol. Correct.

Wait, alternatively, maybe the limiting reactant is Al? No, because 0.0604 mol Al would need 0.0906 mol Cl2, but we have only 0.0752 mol Cl2, so Cl2 is limiting. So theoretical yield is from Cl2.

Wait, let's recalculate theoretical yield using Al: if Al were limiting, 0.0604 mol Al would produce 0.0604 mol AlCl3, mass=0.0604×133.33≈8.05 g. But actual yield is 4.52, which is less than 8.05, but Cl2 is limiting, so theoretical yield must be less than 8.05. So 6.68 g is less than 8.05, so that's correct.

So percent yield = (4.52 / 6.68) × 100 ≈ 67.7%. Wait, but let's check with precise numbers:

Moles of Cl2: 863 mmHg = 863/760 atm = 1.1355263 atm.
V=1.60 L, T=294.15 K.
n = PV/RT = (1.1355263 * 1.60) / (0.082057 * 294.15)
Calculate numerator: 1.1355263 * 1.60 = 1.816842
Denominator: 0.082057 * 294.15 ≈ 24.137
n ≈ 1.816842 / 24.137 ≈ 0.07527 mol Cl2.

Theoretical moles of AlCl3: (2/3)*0.07527 = 0.05018 mol.

Theoretical mass: 0.05018 mol * 133.333 g/mol = 0.05018 * 133.333 ≈ 6.69 g.

Actual yield: 4.52 g.

Percent yield: (4.52 / 6.69) * 100 ≈ 67.6%.

So approximately 67.6% to 67.7%.

Wait, but maybe I made a mistake in the balanced equation? No, 2 Al + 3 Cl2 → 2 AlCl3 is correct.

Alternatively, maybe the limiting reactant is Al? Let's check: 0.0604 mol Al, 0.07527 mol Cl2.

Moles of Cl2 needed for Al: (3/2)*0.0604 = 0.0906 mol. We have 0.07527, which is less, so Cl2 is limiting. Correct.

So percent yield is (4.52 / 6.69) * 100 ≈ 67.6%, which rounds to 67.6% or 67.7%.

Wait, but let's check the actual yield: 4.52 g. Let's calculate theoretical yield from Al: 0.0604 mol Al → 0.0604 mol AlCl3 (since 2 Al → 2 AlCl3). Mass: 0.0604 * 133.333 ≈ 8.05 g. But since Cl2 is limiting, theoretical yield is lower, so 6.69 g. So percent yield is (4.52 / 6.69) * 100 ≈ 67.6%.

So the balanced equation is $2\ce{Al}(s) + 3\ce{Cl2}(g)
ightarrow 2\ce{AlCl3}(s)$, and percent yield is approximately 67.6% (or 67-68%).

Answer:

Balanced equation: $\boldsymbol{2\ce{Al}(s) + 3\ce{Cl2}(g)
ightarrow 2\ce{AlCl3}(s)}$
Percent yield: $\boldsymbol{67.6\%}$ (or ~68%)