# Question 6e99b

Apr 11, 2017

WARNING! Long answer! The alloy is 18.4 % Al and 81.6 % Mg.

#### Explanation:

The chemical equations are:

${\text{2Al" + "6HCl" → "2AlCl"_3 + "3H}}_{2}$
${\text{Mg" + "2HCl" → "MgCl"_2 + "H}}_{2}$

Our first task is to calculate the moles of hydrogen. For this, we can use the Ideal Gas Law:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this formula to get

$n = \frac{p V}{R T}$

In this problem,

p = 684 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.900 atm"
$V = \text{1200 mL" = "1.200 L}$
$R = \text{0.082 16"color(white)(l) "L·atm·K"^"-1""mol"^"-1}$
$T = \text{27 °C" = "(27 + 273.15) K" = "300.15 K}$

n = (0.900 color(red)(cancel(color(black)("atm"))) × 1.200 color(red)(cancel(color(black)("L"))))/("0.082 16" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 300.15 color(red)(cancel(color(black)("K")))) = "0.043 80 mol"

${A}_{\textrm{r}} \left(\text{Al") = 26.98; A_text(r)("Mg}\right) = 24.31$

Let $x \textcolor{w h i t e}{m l} = \text{mass of Al}$; then
$\left(1 - x\right) = \text{mass of Mg}$

(1) Calculate the moles of $\text{Al}$

$\text{Moles of Al" = x color(red)(cancel(color(black)("g Al"))) × "1 mol Al"/(26.98 color(red)(cancel(color(black)("g Al")))) = "0.037 06"x color(white)(l)"mol Al}$

(2) Calculate the moles of ${\text{H}}_{2}$ from $\text{Al}$

$\text{Moles of H"_2 = "0.037 06" x color(red)(cancel(color(black)("mol Al"))) × "3 mol H"_2/(2 color(red)(cancel(color(black)("mol Al")))) = "0.055 60 mol H}$

(3) Calculate the moles of $\text{Mg}$

$\text{Moles of Mg" = (1-x) color(red)(cancel(color(black)("g Mg"))) × "1 mol Mg"/(24.31 color(red)(cancel(color(black)("g Mg")))) = "0.041 14 mol Mg}$

(4) Calculate the moles of ${\text{H}}_{2}$ from $\text{Mg}$

$\text{Moles of H"_2 = "0.041 14" "mol Mg" × "1 mol H"_2/(1 "mol Mg") = "0.041 14 mol H"_2}$

${\text{Moles of H"_2color(white)(l) "from Al + moles of H"_2color(white)(l) "from Mg = total moles of H}}_{2}$

$\textcolor{w h i t e}{m m l l} \text{0.055 60"xcolor(white)(mml) + color(white)(ml)"0.011 44"(1-x)color(white)(m) =color(white)(mml)0.043 80}$

$\textcolor{w h i t e}{m m l l} \text{0.055 60"xcolor(white)(mml) + "0.041 14" - "0.041 14"x = color(white)(mm)"0.043 80}$

$\textcolor{w h i t e}{m m m m m m m m m m m m m m m m l l} \text{0.014 46"x = "0.002 66}$

$x = \text{0.002 66"/"0.014 46} = 0.1840$

$\text{Mass of Al= 0.1840 g}$

$\text{Mass of Mg" = 1-x = "(1 - 0.1840) g" = "0.8160 g}$

"% Al" = (0.1840 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("g")))) × 100 % = 18.4 %

"% Mg" = (0.8160 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("g")))) × 100 % = 81.6 %#

The alloy is 18.4 % Al and 81.6 % Mg.

Check:

${\text{H"_2 color(white)(l)"from Al" = 0.1840 color(red)(cancel(color(black)("g Al"))) × (1 color(red)(cancel(color(black)("mol Al"))))/(26.98 color(red)(cancel(color(black)("g Al")))) × "3 mol H"_2/(2 color(red)(cancel(color(black)("mol Al")))) = "0.010 23 mol H}}_{2}$

${\text{H"_2 color(white)(l)"from Mg" = 0.8160 color(red)(cancel(color(black)("g Mg"))) × (1 color(red)(cancel(color(black)("mol Mg"))))/(24.31 color(red)(cancel(color(black)("g Mgl")))) × "1 mol H"_2/(1 color(red)(cancel(color(black)("mol Mg")))) = "0.033 57 mol H}}_{2}$

$\textcolor{w h i t e}{m m m m m m} {\text{Total moles of H"_2= "0.04 380 mol H}}_{2}$

It checks!