The volume of 10M hydrochloric acid is correct,
As the molecular weight of CaCl2 is 110.98 g/mole
you theoretically need 83.235 g anhydrous calcium chloride,
or 0.75 x (110.98 + 6x18.01528) = 164.3 g if calcium chloride hexahydrate.
The actual hydration level will be unknown, from zero to more than six.
So if you want to work stoichiometrically you need to dehydrate your CaCl2 before weighing.
P.S. With HCl gas you do need a suck-back trap.
[Edited on 7-11-2018 by Sulaiman] |