How many grams of K

Question

How many grams of K₂CO₃ would be needed to prepare 5.00 g of KCl?

Answer 1

K₂CO₃ M.W. = 138.205 g/mol

KCl M.W. = 74.551 g/mol

(5 g KCl) * (mol KCl/74.551 g KCl) * (1 mol K₂CO₃ / 2 mol KCl) *(138.205 g K₂CO₃ /mol K₂CO₃) = ? g K₂CO₃

(5 /74.551)*(1/2 )*(138.205 g K₂CO₃ ) = 4.63 g K₂CO₃

As an aside...

K₂CO₃ + H₂O = KOH + KHCO₃^-

KHCO₃^- + H₂O = KOH + H₂CO₃

so,

K₂CO₃ (aq) + 2 H₂O = 2 KOH (aq) + H₂CO₃ (aq)

So,

Potentially reaction with hydrochloric acid, KOH + HCl = KCl + H₂O, this will produce KCl.

Boiling off the water also drives out CO₂

KCl + H₂O + H₂CO₃ (aq) = KCl + 2H₂O + CO₂ (g)