Question

Whatmassofiron(III)hydroxideprecipitatecanbeproduced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide? ​

Answer 1

volume of Fe(NO3)3, V = 75.0 mL

= 7.5*10^-2 L

number of mol of Fe(NO3)3,

n = Molarity * Volume

= 0.105*0.075

= 7.875*10^-3 mol

volume of NaOH , V = 125 mL

= 0.125 L

number of mol of NaOH,

n = Molarity * Volume

= 0.15*0.125

= 1.875*10^-2 mol

Balanced chemical equation is:

Fe(NO3)3 + 3 NaOH ---> Fe(OH)3 + 3 H2O

1 mol of Fe(NO3)3 reacts with 3 mol of NaOH

for 7.875*10^-3 mol of Fe(NO3)3, 2.36*10^-2 mol of NaOH is required

But we have 1.875*10^-2 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of Fe(OH)3,

MM = 1*MM(Fe) + 3*MM(O) + 3*MM(H)

= 1*55.85 + 3*16.0 + 3*1.008

= 106.874 g/mol

According to balanced equation

mol of Fe(OH)3 formed = (1/3)* moles of NaOH

= (1/3)* 1.875*10^-2

= 6.25*10^-3 mol

mass of Fe(OH)3 = number of mol * molar mass

= 6.25*10^-3*1.069*10^2

= 0.668 g

Answer: 0.668 g