Final answer:
To obtain 198.8 mol of deuterium with a deuterium abundance of 0.01500% and 80% recovery, 14,925.88 kg of water must be processed.
Step-by-step explanation:
To determine how many kilograms of water are needed to obtain 198.8 mol of deuterium when deuterium is 0.01500% (by number) of natural hydrogen, and assuming an 80% recovery rate, we need to calculate the total moles of hydrogen atoms in the water that can yield the required moles of deuterium.
First, let's calculate the total number of moles of hydrogen that contain 198.8 mol of deuterium:
198.8 mol / 0.00015 (deuterium abundance) = 1,325,333.33 mol of hydrogen
Since each water molecule has two hydrogen atoms, we need half the number of moles of water for that many moles of hydrogen:
1,325,333.33 mol H / 2 = 662,666.67 mol of water
Water has a molar mass of 18.015 g/mol, so to find the mass:
662,666.67 mol × 18.015 g/mol = 11,940,701.67 g of water
Considering an 80% recovery rate:
11,940,701.67 g / 0.80 = 14,925,877.09 g
Converting to kilograms:
14,925,877.09 g ÷ 1000 = 14,925.88 kg
Therefore, 14,925.88 kg of water must be processed to obtain 198.8 mol of deuterium.