You'd need "940 g"940 g of zinc metal to react completely with that particular "HCl"HCl solution.
The starting point will be your balanced chemical equation
Zn_((s)) + 2HCl_((aq)) -> ZnCl_(2(aq)) + H_(2(g))Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
Now look at the "1:2"1:2 mole ratio that exists between zinc and hydrochloric acid; every mole of zinc needs 2 moles of hydrochloric acid for the reaction to take place.
This means that you can determine the number of moles of zinc that react by calculating the number of moles of hydrochloric acid that took part in the reaction.
You can do this by using the molarity of the "HCl"HCl solution
C = n/V => n_("HCl") = C * V = "3.5 M" * "8.2 L" = "28.7 moles"C=nV⇒nHCl=C⋅V=3.5 M⋅8.2 L=28.7 moles "HCl"HCl
Using the mole ratio will get you
"28.7 moles HCl" * "1 mole zinc"/"2 moles HCl" = "14.35 moles zinc"28.7 moles HCl⋅1 mole zinc2 moles HCl=14.35 moles zinc
Calculate the mass of zinc by using its molar mass
"14.35 moles" * "65.4 g"/"1 mole" = "938.5 g zinc"14.35 moles⋅65.4 g1 mole=938.5 g zinc
Rounded to two sig figs, the answer will be
m_("zinc") = "940 g"mzinc=940 g
