# Answer to Question #20234 in Physical Chemistry 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.

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

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

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}isM(Fe

_{2}(SO_{4})_{3}) = 2*56+3*96 = 400 g/molThe molar mass of (NH

_{4})_{2}SO_{4}isM((NH

_{4})_{2}SO_{4}) = 18*2 + 96 = 132 g/molThe amount of Fe

_{2}(SO_{4})_{3}isn(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}isn((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}ism((NH

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

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}isC((NH

_{4})_{2}SO_{4}) = n((NH_{4})_{2}SO_{4}) / V(solution) = 0.7125 mol / 0.5 L = 1.425 M (mol/L)Need a fast expert's response?

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