What mass of hydrogen chloride is necessary to neutralize a 75.1*g75.1⋅g mass of calcium hydroxide?

1 Answer
anor277
Feb 28, 2017

Approx. 74*g74⋅g..........

Explanation:

We need (i) a stoichiometric equation:

Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)Ca(OH)2(s)+2HCl(aq)→CaCl2(aq)+2H2O(l).

And (ii), equivalent quantities of calcium hydroxide:

"Moles of calcium hydroxide"=(75.1*g)/(74.09*g*mol^-1)=1.01*molMoles of calcium hydroxide=75.1⋅g74.09⋅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*g2×1.01⋅mol×36.46⋅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"34% conc. hydrochloric acid, which has an approx. molarity of 10.9*mol*L^-110.9⋅mol⋅L−1, what is the volume of hydrochloric acid required for equivalence?

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