Question

A student conducting this experiment weighs out 2.80 g of CoCl₂.6H₂O.

For this part you will be calculating the amount of sodium carbonate required for the complete precipitation of cobalt(II) carbonate.

Calculate the moles of CoCl₂.6H₂O used in this reaction.

Remember to keep all of your signifciant figures on your calculator until the end of the calculation and then round to three significant figures.

Answer 1

The balanced equation for precipitation of  cobalt(II) carbonate is

CoCl₂ .6H₂O(aq) + Na₂CO₃ (aq) 2NaCl(aq) + CoCO3 (s) + .6H₂O(l)

Mass of CoCl₂ .6H₂O used = 2.80 g

Molar mass of CoCl₂ .6H₂O =237.9309 g/mol

No. of moles of CoCl₂ .6H₂O used = given mass / molar mass = ( 2.80 g )/ (237.9309 g/mol)

=0.011768 moles

From the balanced chemical equation, we find that 1 mole of  CoCl₂ .6H₂O requires 1 mole of Na₂CO₃ for complete reaction.

Therefore, 0.011768 moles of  CoCl₂ .6H₂O will require 0.011768 moles of  Na₂CO₃ for complete reaction

Molar mass of Na₂CO₃ = 105.987 g/ mol

Mass of 0.011768 moles of  Na₂CO₃ = (105.987 g/ mol)* (0.011768 moles) = 1.2473 g = 1.25 g

Mass of  Na₂CO₃ required for complete precipitation of  cobalt(II) carbonate is 1.25 g