Question #a22ae

1 Answer
Nov 13, 2016

#"3.7 M"#


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

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

Notice that your reaction consumes #color(red)(2)# moles of hydrochloric acid for every mole of zinc metal present in the coin.

Use the molar mass of zinc to determine how many moles you have in your sample

#3.0 color(red)(cancel(color(black)("g"))) * "1 mole Zn"/(65.38color(red)(cancel(color(black)("g")))) = "0.04589 moles Zn"#

Use the aforementioned mole ratio to find the number of moles of hydrochloric acid needed to ensure that all the moles of zinc metal get converted to zinc cations

#0.04589 color(red)(cancel(color(black)("moles Zn"))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole Zn")))) = "0.09178 moles HCl"#

Now, focus on finding the number of moles of hydrochloric acid present in your initial solution. A #"6.0-M"# hydrochloric acid solution will contain #6.0# moles of hydrochloric acid per liter of solution, i.e. per #10^3"mL"#.

This means that your initial solution contains

#40.0 color(red)(cancel(color(black)("mL solution"))) * "6.0 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.240 moles HCl"#

Your reaction will consume #0.09178# moles of hydrochloric acid, which means that the resulting solution will contain

#overbrace("0.240 moles HCl")^(color(blue)("what you start with")) - overbrace("0.09178 moles HCl")^(color(purple)("what is consumed")) = "0.1482 moles HCl"#

Now all you have to do is use the known volume of the solution to figure out its new molarity. Determine how many moles you'd get per liter of the resulting solution

#10^3 color(red)(cancel(color(black)("mL"))) * "0.1482 moles HCl"/(40.0color(red)(cancel(color(black)("mL")))) = "3.706 moles HCl"#

This means that after the reaction is complete, the molarity of the solution will be

#color(darkgreen)(ul(color(black)("3.7 mol L"^(-1) = "3.7 M")))#

The answer is rounded to two sig figs.