Question

How many kilograms of water must be processed to obtain \(2.0 \mathrm{~L}\) of \(\mathrm{D}_{2}\) at \(25^{\circ} \mathrm{C}\) and 0.90 atm pressure? Assume that deuterium abundance is 0.015 percent and that recovery is 80 percent.

Short answer

Approximately 21.91 kg of water is needed.

Step-by-step solution

Understand the Problem

We need to determine the amount of water in kilograms required to produce 2.0 L of deuterium gas (\(\mathrm{D}_2\)) under the given conditions. We have these calculations based on abundance and recovery rate.

Convert the Desired Volume to Moles

First, we use the ideal gas law to determine the moles of \(\mathrm{D}_2\) needed. The ideal gas law is \(PV=nRT\), where:- \(P = 0.90\, \text{atm}\)- \(V = 2.0\, \text{L}\)- \(R = 0.0821\, \text{L atm K}^{-1}\, \text{mol}^{-1}\)- \(T = (25 + 273)\, \text{K} = 298\, \text{K}\)Using the equation, \(n = \frac{PV}{RT} = \frac{0.90 \times 2.0}{0.0821 \times 298}\approx 0.073\, \text{mol}\).

Calculate Moles of Water Required

Since deuterium is only 0.015% of hydrogen, and since each water molecule has 2 hydrogen atoms, it takes \(100/0.015\times 2\approx 13333\)n moles of water to get 1 mole of \(\mathrm{D}_2\) gas. Therefore, for \(0.073\) moles of \(\mathrm{D}_2\), we need:\[13333 \times 0.073 = 973.309\, \text{mol}\]of water.

Adjust for Recovery Rate

Considering that only 80% of deuterium is recovered, the actual moles required are increased by a factor of \(\frac{1}{0.80}\):\[\frac{973.309}{0.80} \approx 1216.64\, \text{mol}\].

Convert Moles of Water to Kilograms

Since the molar mass of \(\mathrm{H}_2\mathrm{O}\) is approximately \(18.015\, \text{g/mol}\), we convert moles to mass:\[1216.64 \times 18.015 = 21,913.26\, \text{g}\]Convert grams to kilograms:\[\frac{21,913.26}{1000} \approx 21.91\, \text{kg}\].

Key concepts

Deuterium Recovery

Deuterium, often referred to as heavy hydrogen, is an isotope of hydrogen with an extra neutron, making it twice as heavy. Recovering deuterium from water involves processes like distillation or electrolysis, and one must consider its natural abundance and recovery efficiency. In our exercise, the abundance is given as 0.015%, meaning for every 10,000 hydrogen atoms, there is roughly 1.5 deuterium atoms. This extremely low natural abundance necessitates processing large amounts of water to obtain even a small quantity of deuterium.

• Most processes for deuterium recovery have a recovery rate percentage, which represents the efficiency of the separation process.
• Here, the recovery rate is 80%, meaning only 80% of available deuterium in the water sample is actually recovered.

To account for this in calculations, the moles required must be adjusted by dividing by the recovery percentage (as a decimal), ensuring you factor in the efficiency of your recovery process.

Molar Mass Calculation

Understanding molar mass is essential for any chemical calculation. Molar mass represents the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). For water (H₂O), the calculation is straightforward by adding the atomic masses:

• Hydrogen (H): approximately 1.008 g/mol
• Oxygen (O): approximately 16.00 g/mol

So, the molar mass of H₂O equals about 18.015 g/mol in total. This value is critical when converting between moles and grams. In chemical calculations like this exercise, after determining the moles of water needed, you multiply the number of moles by this molar mass to obtain the mass:

• Example: 1216.64 mol of water × 18.015 g/mol equals 21,913.26 grams

Remember to convert grams to kilograms by dividing by 1000, which yields the final mass in kilograms for the calculations.

Chemical Calculations

Chemical calculations often involve the use of the ideal gas law to relate volume, pressure, and temperature to the amount of gas (in moles). The ideal gas law is expressed as:\[PV = nRT\]where:

• \(P\) is pressure in atmospheres (atm)
• \(V\) is volume in liters (L)
• \(n\) is number of moles
• \(R\) is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹)
• \(T\) is temperature in Kelvin (K), calculated by adding 273 to Celsius degrees

Once you calculate the number of moles, you can apply stoichiometry to determine the amount of reactant needed or product formed. In our exercise, deriving the moles of deuterium using the ideal gas law helps determine the moles of water required. The stoichiometric relationships, adjusted for deuterium's low abundance, allow accurate calculations for large-scale chemical processes.