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

a)

Fe_{2}(SO_{4})_{3} + 6NH_{3} + 6H_{2}O = 2Fe(OH)_{3} + 3(NH_{4})_{2}SO_{4}

b)

The molar mass of Fe_{2}(SO_{4})_{3} is

M(Fe_{2}(SO_{4})_{3}) = 2*56+3*96 = 400 g/mol

The molar mass of (NH_{4})_{2}SO_{4} is

M((NH_{4})_{2}SO_{4}) = 18*2 + 96 = 132 g/mol

The amount of Fe_{2}(SO_{4})_{3} is

n(Fe_{2}(SO_{4})_{3}) = m(Fe_{2}(SO_{4})_{3}) / M(Fe_{2}(SO_{4})_{3}) = 95 g / 400g/mol = 0.2375 mol According to reaction above the amount of (NH_{4})_{2}SO_{4} is

n((NH_{4})_{2}SO_{4}) = 3 * n(Fe_{2}(SO_{4})_{3}) = 3*0.2375 mol =0.7125 mol The mass of (NH_{4})_{2}SO_{4} is

m((NH_{4})_{2}SO4) = n((NH_{4})_{2}SO_{4}) * M((NH_{4})_{2}SO_{4}) = 0.7125 mol * 132 g/mol = 94.05 g

c)

The amount of (NH_{4})_{2}SO_{4} is 0.7125 mol, the total volume of the solution is 500mL. Thus, the molar concentration of (NH_{4})_{2}SO_{4} is

C((NH_{4})_{2}SO_{4}) = n((NH_{4})_{2}SO_{4}) / V(solution) = 0.7125 mol / 0.5 L = 1.425 M (mol/L)

Fe

b)

The molar mass of Fe

M(Fe

The molar mass of (NH

M((NH

The amount of Fe

n(Fe

n((NH

m((NH

c)

The amount of (NH

C((NH

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