The chemical equations are:
#"2Al" + "6HCl" → "2AlCl"_3 + "3H"_2#
#"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:
#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#
We can rearrange this formula to get
#n = (pV)/(RT)#
In this problem,
#p = 684 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.900 atm"#
#V = "1200 mL" = "1.200 L"#
#R = "0.082 16"color(white)(l) "L·atm·K"^"-1""mol"^"-1"#
#T = "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_text(r)("Al") = 26.98; A_text(r)("Mg") = 24.31#
Let #xcolor(white)(ml) = "mass of Al"#; then
#(1-x) = "mass of Mg"#
(1) Calculate the moles of #"Al"#
#"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 #"H"_2# from #"Al"#
#"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 #"Mg"#
#"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 #"H"_2# from #"Mg"#
#"Moles of H"_2 = "0.041 14" "mol Mg" × "1 mol H"_2/(1 "mol Mg") = "0.041 14 mol H"_2"#
#"Moles of H"_2color(white)(l) "from Al + moles of H"_2color(white)(l) "from Mg = total moles of H"_2#
#color(white)(mmll)"0.055 60"xcolor(white)(mml) + color(white)(ml)"0.011 44"(1-x)color(white)(m) =color(white)(mml)0.043 80"#
#color(white)(mmll)"0.055 60"xcolor(white)(mml) + "0.041 14" - "0.041 14"x = color(white)(mm)"0.043 80"#
#color(white)(mmmmmmmmmmmmmmmmll)"0.014 46"x = "0.002 66"#
#x = "0.002 66"/"0.014 46" = 0.1840#
∴ #"Mass of Al= 0.1840 g"#
#"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:
#"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#
#"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#
#color(white)(mmmmmm)"Total moles of H"_2= "0.04 380 mol H"_2#
It checks!
