What volume of 3.05 M #HCl(aq)# will react with 25.0 g #Zn(s)# in the reaction #Zn(s) + 2HCl(aq) -> ZnCl_2(aq) + H_2(g)#?

Start by writing down the balanced chemical equation that describes this single replacement reaction

#"Zn"_ ((s)) + color(red)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H"_(2(g)) uarr#

Now, notice that the two reactants react in a #1:color(red)(2)# mole ratio. This tells you that the reaction will always consume twice as many moles of hydrochloric acid than moles of zinc metal.

As you know, a solution's molarity tells you how many moles of solute, which in your case will be hydrochloric acid, #"HCl"#, are present per liter of solution.

The hydrochloric acid solution is said to have a molarity of #"3.05 mol L"^(-1)#. This means that you get #3.05# moles of hydrochloric acid for every #"1 L"# of solution. Keep this in mind for later.

You know that the reaction must consume #"25.0 g"# of zinc metal. You can convert this mass to moles by using zinc's molar mass

#25.0 color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38color(red)(cancel(color(black)("g")))) = "0.3824 moles Zn"#

In order for this many moles of zinc to react, you'd need

#0.3824 color(red)(cancel(color(black)("moles Zn"))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole Zn")))) = "0.7648 moles HCl"#

Since you know that #"1 L"# of your hydrochloric acid solution contains #3.05# moles of hydrochloric acid, ti follows that this many moles would be present in

#0.7648 color(red)(cancel(color(black)("moles HCl"))) * overbrace("1 L solution"/(3.05 color(red)(cancel(color(black)("moles HCl")))))^(color(purple)(" a molarity of 3.05 mol L"^(-1))) = "0.2508 L solution"#

Rounded to three sig figs and expressed in milliliters, the answer will be

#"volume of HCl solution" = color(green)(|bar(ul(color(white)(a/a)"251 mL"color(white)(a/a)|)))#

Note that you have

#"1 L" = 10^3"mL"#

SIDE NOTE The hydrogen gas produced by the reaction will bubble out of solution.