The formula for Mohr's salt is #("NH"_4)_2"Fe"("SO"_4)_2·6"H"_2"O"#.
To avoid having to write complicated formulas, let's use #"M"# to represent Mohr's salt and #"A"# to represent ammonium sulfate.
The equations for the reaction with #"BaCl"_2# are
#M_text(r):color(white)(m)123.14color(white)(mmmmmmmll)233.39#
#color(white)(mm)("NH"_4)_2"SO"_4 + "BaCl"_2 → "BaSO"_4 + "2NH"_4"Cl"#
and
#M_text(r):color(white)(mmmmmm)392.14color(white)(mmmmmmmmmll)233.39#
#color(white)(mm)("NH"_4)_2"Fe"("SO"_4)_2·6"H"_2"O" + "2BaCl"_2 → "2BaSO"_4 + "4NH"_4"Cl" + "FeCl"_2 + 6"H"_2"O"#
Calculate the mass of #"BaSO"_4# from #"A"#
Let #x# be the mass of #"A"#. Then #0.5 - x# is the mass of #"M"#.
#"Mass of BaSO"_4 = x color(red)(cancel(color(black)("g A"))) × (1 color(red)(cancel(color(black)("mol A"))))/(123.14 color(red)(cancel(color(black)("g A")))) × (1 color(red)(cancel(color(black)("mol BaSO"_4))))/(1 "mol A") × ("233.99 g BaSO"_4)/(1 color(red)(cancel(color(black)("mol BaSO"_4)))) = 1.9002 x color(white)(l)"g BaSO"_4 #
Calculate the mass of #"BaSO"_4# from #"M"#
#"Mass of BaSO"_4#
#= (0.5 - x) color(red)(cancel(color(black)("g M"))) × (1 color(red)(cancel(color(black)("mol M"))))/(392.14 color(red)(cancel(color(black)("g M")))) × (2 color(red)(cancel(color(black)("mol BaSO"_4))))/(1 "mol M") × ("233.99 g BaSO"_4)/(1 color(red)(cancel(color(black)("mol BaSO"_4)))) = 1.1934(0.5-x) color(white)(l)"g BaSO"_4 = ("0.5967 - "1.1934x) "g BaSO"_4#
Calculate the masses of #"A"# and #"M"#
The total mass of #"BaSO"_4# is 0.75 g.
∴ #1.9002x + "0.5967 - 1.1934"x = 0.75#
#0.7068x = "0.75 - 0.5967" = 0.153#
#x = 0.153/0.7068 = 0.22#
The mass of #"A"# is 0.22 g.
The total mass of #"A + M"# is 0.5 g.
∴ The mass of #"M"# is (0.5 - 0.22) g = 0.28 g
Calculate the % composition of the mixture
#"% A" = (0.22 color(red)(cancel(color(black)("g"))))/(0.5 color(red)(cancel(color(black)("g")))) × 100 % = 43 % #
#"% M" = "(100 - 43) %" = 57 %"#
Calculate the mass of #"Fe"_2"O"_3#
The #"Fe"_2"O"_3# comes from the Mohr's salt.
(a) Calculate the mass of #"M"# in the sample
#"Mass of M in sample" = 0.2 color(red)(cancel(color(black)("g sample"))) × "57 g M"/(100 color(red)(cancel(color(black)("g Sample")))) = "0.114 g M"#
(b) Write the partial equation for the pyrolysis reaction
#M_text(r):color(white)(mmmll)392.14color(white)(mmmmmmm)159.69#
#color(white)(mm)2("NH"_4)_2"Fe"("SO"_4)_2·6"H"_2"O" → "Fe"_2"O"_3#
(c) Calculate the mass of #"Fe"_2"O"_3#
#"Mass of Fe"_2"O"_3 = 0.114 color(red)(cancel(color(black)("g M"))) × (1 color(red)(cancel(color(black)("mol M"))))/(392.14 color(red)(cancel(color(black)("g M")))) × (1 color(red)(cancel(color(black)("mol Fe"_2"O"_3))))/(2 color(red)(cancel(color(black)("mol M")))) × ("159.69 g Fe"_2"O"_3)/(1 color(red)(cancel(color(black)("mol Fe"_2"O"_3)))) = "0.023 g Fe"_2"O"_3 = "23 mg Fe"_2"O"_3#
