You are out by a multiplier factor of 100.
If you have 42000 L tank containing 120 mg/L (ppm) of Lithium brine solution, Li+ ions, or 0.12 g/L
The tank contains 42000 x 0.12 = 5040 g of Li ions
To convert the Li ions to LCE you need 2 mole of Li ions per 1 mole of Li2CO3
Since 5040 g of Li is 5040/6.94 = 726.2 mole of Lithium which will produce 363.1 moles of Li2CO3
Which is 363.1 x 73.89 = 26830 g of Li2CO3 or 26.8 kg LCE assuming 100% conversion
Ann: Lithium Drilling Approval at Prairie, page-115
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