A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?

The balanced chemical equation for the electrolysis of molten barium chloride is: \[ \mathrm{BaCl_{2}(l) \rightarrow Ba(s) + Cl_{2}(g)} \] For every mole of barium produced, two moles of electrons are consumed.

To convert the given mass of barium to moles, we will need the molar mass of barium (Ba). From the periodic table, the molar mass of barium is: \[ \mathrm{Molar~mass~of~Ba = 137.33~g/mol} \]

Now, we need to convert the given amount of barium (\(1.00 \times 10^{3} \mathrm{~kg}\)) to moles. To do this, we'll first convert the given mass from kg to grams, then use the molar mass of barium to calculate the moles. \[ \mathrm{Amount~of~Ba = \frac{1.00 \times 10^{3} kg \times 10^{3} \frac{g}{kg}}{137.33 \frac{g}{mol}} \] \[ \mathrm{Amount~of~Ba = 7.28 \times 10^{3}~mol} \]

Now that we know the moles of barium produced, we can find how many moles of electrons are consumed using stoichiometry from the balanced chemical equation. Since two moles of electrons are consumed for every mole of barium produced, we can multiply the moles of barium by 2. \[ \mathrm{Moles~of~electrons = Amount~of~Ba \times 2} \] \[ \mathrm{Moles~of~electrons = 7.28 \times 10^{3}~mol \times 2} \] \[ \mathrm{Moles~of~electrons = 1.46 \times 10^{4}~mol} \]

With the calculated moles of electrons, we can now use Faraday's law of electrolysis to find the required current. Faraday's law states that charge \((Q)\) needed to produce a substance in an electrolysis process is: \[ Q = n \times F \] Where \(n\) is the moles of electrons consumed, and \(F\) is the Faraday constant, which is approximately \(96485 \mathrm{C/mol}\). Once we have the total charge, we can find the current \((I)\) using the relationship \(Q = It\), where \(t\) is the time in seconds. First, let's calculate the total charge (Q): \[ \mathrm{Q = n \times F = 1.46 \times 10^{4}~mol \times 96485 \frac{C}{mol}} \] \[ \mathrm{Q = 1.41 \times 10^{9}~C} \] Now, let's convert the given time into seconds: \[ \mathrm{Time~(in~seconds) = 4.00~h \times 3600 \frac{s}{h} = 1.44 \times 10^{4}~s} \] Finally, calculate the current required: \[ \mathrm{I = \frac{Q}{t} = \frac{1.41 \times 10^{9}~C}{1.44 \times 10^{4}~s}} \] \[ \mathrm{I = 9.79 \times 10^{4}~A} \] Hence, the current needed to produce \(1.00 \times 10^{3} \mathrm{~kg}\) of barium from the electrolysis of molten barium chloride in 4 hours is approximately \(9.79 \times 10^{4} \mathrm{~A}\).