Question

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?

Answer 1

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.