Question

What mass of hydrogen chloride is necessary to neutralize a `75.1*g` mass of calcium hydroxide?

Answer 1

Approx. `74*g`..........

Explanation:

We need (i) a stoichiometric equation:

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

And (ii), equivalent quantities of calcium hydroxide:

`"Moles of calcium hydroxide"=(75.1*g)/(74.09*g*mol^-1)=1.01*mol`.

And for an equivalent quantity of hydrochloric acid, we thus need `2xx1.01*molxx36.46*g*mol^-1=73.6*g`.

Why did I double the molar quantity of calcium hydroxide?

Given that we would normally use `"34% conc. hydrochloric acid"`, which has an approx. molarity of `10.9*mol*L^-1`, what is the volume of hydrochloric acid required for equivalence?