# Question 4492b

Apr 23, 2016

$\text{1300 mL}$

#### Explanation:

The idea here is that you need to use the balanced chemical equation that describes this single replacement reaction to determine how many grams of hydrochloric acid, $\text{HCl}$, are needed in order to produce that much hydrogen gas, ${\text{H}}_{2}$.

${\text{Mg"_ ((s)) + color(red)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + "H}}_{2 \left(g\right)} \uparrow$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between hydrochloric acid and hydrogen. This tells you that the reaction will produce $1$ mole of hydrogen gas for every $\textcolor{red}{2}$ moles of hydrochloric acid that take part in the reaction.

Use hydrogen gas' molar mass to determine how many moles you have in that $\text{10-g}$ sample

10 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "4.96 moles H"_2

This means that the reaction must have consumed

4.96 color(red)(cancel(color(black)("moles H"_2))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole H"_2)))) = "9.92 moles HCl"

Now use hydrochloric acid's molar mass to determine how many grams would contain this many moles

9.92 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "361.7 g HCl"

Now, you know that you're working with a solution that $\text{27% w/v}$ hydrochloric acid. This means that every $\text{100 mL}$ of this solution will contain $\text{27 g}$ of hydrochloric acid.

As it turns out, you don't need to use the density of the solution to find the volume that would contain $\text{361.7 g}$ of hydrochloric acid because you can get that directly from the solution's the mass by volume percent concentration.

So, if get $\text{27 g}$ of hydrochloric acid per $\text{100 mL}$ of this solution, it follows that you will need

361.7 color(red)(cancel(color(black)("g HCl"))) * overbrace("100 mL solution"/(27color(red)(cancel(color(black)("g HCl")))))^(color(purple)("= 27% w/v")) = "1340 mL"

of solution to provide $\text{361.7 g}$ of hydrochloric acid to the reaction.

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig figf for the mass of hydrogen gas produced by the reaction

"volume of 27% w/v HCl solution" = color(green)(|bar(ul(color(white)(a/a)"1300 mL"color(white)(a/a)|)))#