How many grams of lithium carbonate are needed to make a concentration of 12.7M in a 5L solution

The answer to your question is  4699 g

Explanation:

Data

mass of Li₂CO₃ = ?

concentration = 12.7 M

volume = 5 l

Process

To solve this problem use the formula of Molarity and solve it for moles.

1.- Write the formula of Molarity

Molarity = moles / volume

-Solve it for moles

moles = Molarity x volume

2.- Substitute the values

moles = 12.7 x 5

3.- Result

moles = 63.5

4.- Calculate the molar mass of Li₂CO₃

Li₂CO₃ = (7 x 2) + (12 x 1) + (16 x 3)

            = 14 + 12 + 48

            = 74 g

5.- Use proportions to calculate the moles

              74 g of Li₂CO₃ 1 mol

                x                     63.5 moles

               x = (63.5 x 74 ) /

               x = 4699 g

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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