Answer to Question #20234 in Physical Chemistry for Makhosini Ndzingane

Question

Balance the following chemical equation:
a) Fe2(SO4)3 + NH3 + H2O = Fe(OH)3 + (NH4)2SO4
b) Using the reaction in (a) how much (NH4)2SO4 in grams would be produced from 95g of Fe2(SO4)3
c) If the total volume of the solution was 500 ml , what would be final concentration of (NH4)2SO4 in moles per L.

Answer 1

a)
Fe₂(SO₄)₃ + 6NH₃ + 6H₂O = 2Fe(OH)₃ + 3(NH₄)₂SO₄
b)
The molar mass of Fe₂(SO₄)₃ is
M(Fe₂(SO₄)₃) = 2*56+3*96 = 400 g/mol
The molar mass of (NH₄)₂SO₄ is
M((NH₄)₂SO₄) = 18*2 + 96 = 132 g/mol
The amount of Fe₂(SO₄)₃ is
n(Fe₂(SO₄)₃) = m(Fe₂(SO₄)₃) / M(Fe₂(SO₄)₃) = 95 g / 400g/mol = 0.2375 mol According to reaction above the amount of (NH₄)₂SO₄ is
n((NH₄)₂SO₄) = 3 * n(Fe₂(SO₄)₃) = 3*0.2375 mol =0.7125 mol The mass of (NH₄)₂SO₄ is
m((NH₄)₂SO4) = n((NH₄)₂SO₄) * M((NH₄)₂SO₄) = 0.7125 mol * 132 g/mol = 94.05 g
c)
The amount of (NH₄)₂SO₄ is 0.7125 mol, the total volume of the solution is 500mL. Thus, the molar concentration of (NH₄)₂SO₄ is
C((NH₄)₂SO₄) = n((NH₄)₂SO₄) / V(solution) = 0.7125 mol / 0.5 L = 1.425 M (mol/L)