An aluminum production cell of the type pictured in Figure \(21-24\) operates at a current of \(1.00 \times 10^{5} \mathrm{A}\) and a voltage of 4.5 V. The cell is \(38 \%\) efficient in using electrical energy to produce chemical change. (The rest of the electrical energy is dissipated as thermal energy in the cell.) (a) What mass of \(\mathrm{Al}\) can be produced by this cell in \(8.00 \mathrm{h} ?\) (b) If the electrical energy required to power this cell is produced by burning coal \((85 \%\) C; heat of combustion of \(C=32.8 \mathrm{kJ} / \mathrm{g}\) ) in a power plant with \(35 \%\) efficiency, what mass of coal must be burned to produce the mass of Al determined in part (a)?

Step by step solution

01

Calculate the power and energy produced by the cell

Using the formula \(P=VI\), where P is power, V is voltage, and I current, the power produced by the cell can be calculated as \(P=1.00 \times 10^{5} \mathrm{A} \times 4.5\mathrm{V} = 450000 \mathrm{W}\) or \(450 \mathrm{KW}\). To calculate the energy produced in 8 hours, use the formula \(E=Pt\), where E is energy, P power and t time (in seconds). Therefore \(E=450 \mathrm{KW} \times 8 \times 3600 \mathrm{s} = 12960000 \mathrm{KJ}\).

02

Calculate the mass of Al produced

Given the efficiency of the cell as 38%, the actual energy used in producing the Al is \(0.38 \times 12960000 \mathrm{KJ} = 4924800 \mathrm{KJ}\). The reaction for the production of Al from Al2O3 is 2Al2O3 → 4Al + 3O2, where \(2 \mathrm{mol} \mathrm{Al2O3} = 4 \mathrm{mol} \(\mathrm{Al} = 393.5 \mathrm{KJ}. Therefore, the number of moles of Al produced is \(4924800 \mathrm{KJ}\) / \(393.5 \mathrm{KJ/mol}\) = \(12511.15 \mathrm{mol} . The mass of Al produced can be found using the molar mass of Al, 27 g/mol: 12511.15 mol x 27g/mol = 337801.5 g or 337.8 kg .

03

Find the mass of Coal required to produce the energy

The total energy required, given the efficiency of the power plant is 35%, is \(E_{total} = 12960000 \mathrm{KJ}\) / \(0.35 = 37028571.43 \mathrm{KJ} . Since the heat of combustion of coal is 32.8 kJ/g, the mass of coal required is \( m = 37028571.43 \mathrm{KJ}\) / \(32.8 \mathrm{KJ/g} = 1129620.04 g or 1129.62 \(\mathrm{Kg}\).

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