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

1 Answer
anor277
Feb 28, 2017

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?

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