# Question dc306

May 16, 2015

You'll be left with 1.82 g of zinc.

$Z {n}_{\left(s\right)} + \textcolor{red}{2} H C {l}_{\left(a q\right)} \to Z n C {l}_{2 \left(a q\right)} + {H}_{2 \left(g\right)}$

Notice the $1 : \textcolor{red}{2}$ mole ratio that exists between zinc and hydrochloric acid; this tells you that, regardless of how many moles of zinc you have, you need twice as many moles of hydrochloric acid for the reaction to take place.

Anything less than that, and hydrochloric acid will be a limiting reagent.

Use zinc's molar mass to determine how many moles you have

6.72cancel("g") * "1 mole Zn"/(65.29cancel("g")) = "0.1028 moles Zn"

Now use the hydrochloric acid solution's molarity to determine how mnay moles of $H C l$ you have

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{H C l} = \text{1.50 M" * 100.0 * 10^(-3)"L" = "0.150 moles HCl}$

This many moles of zinc would have needed

0.1028cancel("moles Zn") * (color(red)(2)"moles HCl")/(1cancel("mole Zn")) = "0.2056 moles HCl"

Since you have less than that, $H C l$ will be a limiting reagent, i.e. it will determine how much zinc actually reacts.

0.150cancel("moles HCl") * "1 mole Zn"/(color(red)(2)cancel("moles HCl")) = "0.075 moles Zn"

Only 0.075 moles of zinc will take part in the rection, the rest will be in excess.

n_("Zn excess") = 0.01028 - 0.075 = "0.0278 moles Zn" $\to$ excess

This is equivalent to having

0.0278cancel("moles Zn") * "65.39 g"/(1cancel("mole Zn")) = "1.818 g"#

That much zinc will remain in solution.

Rounded to three sig figs, the answer will be

${m}_{\text{Zn remaining in solution") = color(green)("1.82 g}}$