Question

a current of 5.25 a is applied to the aluminum sulfate solution for 112 minutes. what mass of al is plated onto the part?
al³⁺ + 3e⁻ → al
mass = ? g al

Explanation:

Step1: Calculate the total charge (Q)

The formula for charge is \( Q = I \times t \), where \( I \) is current in amperes (A) and \( t \) is time in seconds (s). First, convert time from minutes to seconds: \( 112 \, \text{min} \times 60 \, \text{s/min} = 6720 \, \text{s} \). Then, \( Q = 5.25 \, \text{A} \times 6720 \, \text{s} = 35280 \, \text{C} \) (since \( 1 \, \text{A} = 1 \, \text{C/s} \)).

Step2: Determine moles of electrons

Faraday's constant \( F = 96485 \, \text{C/mol e}^- \). Moles of electrons (\( n_{e^-} \)) is \( \frac{Q}{F} \), so \( n_{e^-} = \frac{35280 \, \text{C}}{96485 \, \text{C/mol e}^-} \approx 0.3656 \, \text{mol e}^- \).

Step3: Relate moles of electrons to moles of Al

From the reaction \( \text{Al}^{3+} + 3\text{e}^- ightarrow \text{Al} \), the mole ratio of \( \text{Al} \) to \( \text{e}^- \) is \( 1:3 \). So moles of \( \text{Al} \) (\( n_{\text{Al}} \)) is \( \frac{n_{e^-}}{3} = \frac{0.3656 \, \text{mol e}^-}{3} \approx 0.1219 \, \text{mol Al} \).

Step4: Calculate mass of Al

Molar mass of \( \text{Al} \) is \( 26.98 \, \text{g/mol} \). Mass (\( m \)) is \( n \times M \), so \( m = 0.1219 \, \text{mol} \times 26.98 \, \text{g/mol} \approx 3.29 \, \text{g} \) (more precise calculation: \( 0.3656/3 \times 26.98 \approx (35280 / (96485 \times 3)) \times 26.98 \approx (35280 \times 26.98) / (96485 \times 3) \approx 951854.4 / 289455 \approx 3.29 \, \text{g} \)). Wait, let's recalculate more accurately:

First, \( Q = 5.25 * 112 * 60 = 5.25 * 6720 = 35280 \, \text{C} \)

Moles of e⁻: \( 35280 / 96485 ≈ 0.3656 \)

Moles of Al: \( 0.3656 / 3 ≈ 0.12187 \)

Mass: \( 0.12187 * 26.9815 ≈ 0.12187 * 26.98 ≈ 3.29 \, \text{g} \)? Wait, no, let's do it with exact fractions:

\( \text{Mass} = \frac{I \times t \times M_{\text{Al}}}{3 \times F} \)

Plugging in: \( I = 5.25 \), \( t = 112 \times 60 = 6720 \), \( M_{\text{Al}} = 26.9815 \), \( F = 96485 \)

\( \text{Mass} = \frac{5.25 \times 6720 \times 26.9815}{3 \times 96485} \)

Calculate numerator: \( 5.25 * 6720 = 35280 \); \( 35280 * 26.9815 ≈ 35280 * 26.98 ≈ 35280*20 + 35280*6 + 35280*0.98 = 705600 + 211680 + 34574.4 = 951854.4 \)

Denominator: \( 3 * 96485 = 289455 \)

\( 951854.4 / 289455 ≈ 3.29 \, \text{g} \)? Wait, maybe I made a mistake in calculation. Wait, let's check again:

Wait, 5.25 A * 112 min * 60 s/min = 5.25 * 6720 = 35280 C.

Moles of electrons: 35280 / 96485 ≈ 0.3656 mol e⁻.

Moles of Al: 0.3656 / 3 ≈ 0.1219 mol Al.

Mass: 0.1219 mol * 26.98 g/mol ≈ 0.1219 * 26.98 ≈ 3.29? Wait, no, 0.1219 * 27 ≈ 3.3, but more accurately:

26.9815 * 0.12187 ≈ 26.9815 * 0.1 = 2.69815; 26.9815 * 0.02 = 0.53963; 26.9815 * 0.00187 ≈ 0.0504. Sum: 2.69815 + 0.53963 = 3.23778 + 0.0504 ≈ 3.288 ≈ 3.29 g. Wait, but let's do the calculation with more precision:

35280 / 96485 = let's divide numerator and denominator by 5: 7056 / 19297 ≈ 0.3656.

0.3656 / 3 = 0.12187.

0.12187 * 26.981538 ≈ 0.12187 * 26.9815 ≈ let's compute 26.9815 * 0.1 = 2.69815; 26.9815 * 0.02 = 0.53963; 26.9815 * 0.00187 = 26.9815 * 0.001 + 26.9815 * 0.00087 = 0.0269815 + 0.0234739 = 0.0504554. Then total: 2.69815 + 0.53963 = 3.23778 + 0.0504554 = 3.2882354 ≈ 3.29 g? Wait, but maybe I messed up the ratio. Wait, the reaction is Al³+ + 3e⁻ → Al, so 3 moles of electrons per 1 mole of Al. So moles of Al = moles of e⁻ / 3. That's correct.

Wait, let's use exact formula:

Mass = (I * t * M) / (n * F), where n is the number of electrons per mole of metal (here n=3), M is molar mass of Al (26.98 g/mol), I=5.25 A, t=112*60 s.

So:

I = 5.25, t = 112*60 = 6720, M=26.98, n=3, F=96485.

Mass = (5.25 * 6720 * 26.98) / (3 * 96485)

Calculate numerator: 5.25*6720 = 35280; 35280*26.98 = 35280*(27 - 0.02) = 35280*27 - 35280*0.02 = 952560 - 705.6 = 951854.4

Denominator: 3*96485 = 289455

951854.4 / 289455 ≈ 3.29 g. Wait, but let's check with a calculator: 951854.4 ÷ 289455 ≈ 3.29. Yes.

Answer:

\boxed{3.29} (or more accurately, let's recalculate:

Wait, maybe I made a mistake in the number of electrons. Wait, Al³+ gains 3 electrons, so 3 moles of e⁻ per 1 mole of Al. So the formula is correct.

Wait, another way:

1. Charge: 5.25 A 112 min 60 s/min = 5.25 * 6720 = 35280 C.

1. Moles of electrons: 35280 C / 96485 C/mol e⁻ ≈ 0.3656 mol e⁻.

1. Moles of Al: 0.3656 mol e⁻ * (1 mol Al / 3 mol e⁻) ≈ 0.1219 mol Al.

1. Mass of Al: 0.1219 mol * 26.98 g/mol ≈ 3.29 g.

Yes, so the mass is approximately 3.29 g.