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Answer on Physical Chemistry Question for Makhosini Ndzingane

Question #20234
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.
Expert's answer
Fe2(SO4)3 + 6NH3 + 6H2O = 2Fe(OH)3 + 3(NH4)2SO4
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
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)

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