Question

How many mL of 14.5M lithium carbonate solution must be used to deliver 4.20 g of lithium ion

Answer 1

10.3 g Li) / (6.9410 g Li/mol) x (1 mol Li3PO4 / 3 mol Li) / (0.750 mol/L Li3PO4) = 0.6595 L = 660. mL

Step-by-step explanation: