Question

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)`?

Answer 1

`"251 mL"`

Explanation:

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.