Question

If I combine 15.0 grams of calcium hydroxide with 75.0 mL of 0.500M HCl, how many grams of calcium chloride would be formed?

Answer 1

Approx...`2*g`...

Explanation:

We follow the stoichiometric equation....

`Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O`

`n_(Ca(OH)_2)=(15.0*g)/(74.09*g*mol^-1)=0.203*mol`

And we got with respect to `HCl` a molar quantity of...

`75.0*mLxx0.500*mol*L^-1xx10^-3*L*mL^-1=0.0375*mol`

Clearly the calcium hydroxide is present in molar excess....and at most we can form `(0.0375*mol)/2` of calcium chloride...a mass of `110.98*g*mol^-1xx(0.0375*mol)/2=??*g`..

The question operates under a faulty premise...because the species in solution after addition of the acid would be `Ca(OH)Cl`