How many grams of solid zinc metal will completely react with 100.0 mL of 0.0525 M HCl, if zinc chloride and hydrogen gas are the only products?

1 Answer
anor277
Apr 29, 2017

Approx. #200*mg#.............

Explanation:

The first step is to write the stoichiometric equation, in order to establish the molar equivalence.........

#Zn(s) + 2HCl(aq) rarr ZnCl_2(aq) + H_2(g)uarr#

And then we use the relationship,

#"Concentration"="Moles of solute"/"Volume of solution".#

And thus #"Moles of solute"="concentration"xx"volume of solution"#.

And so here..............

#"Moles of HCl"=100*cancel(mL)xx10^-3cancel(L*mL^-1)xx0.0525*mol*cancel(L^-1)#

#=5.25xx10^-3*mol.#

And thus we require a half equiv of zinc metal, i.e.

#=5.25xx10^-3*molxx1/2xx65.4*g*mol^-1=0.171*g# zinc metal........

But typically we would add a known stoichiometric EXCESS of hydrochloric acid, and either collect the dihydrogen gas evolved, OR back titrate with standardized #NaOH(aq)# solution.

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