Question

Mixing solutions of iron(III) nitrate and sodium hydroxide precipitates the pigment iron(III) hydroxide. What mass of iron(III) hydroxide is produced from 24.1 grams of sodium hydroxide?

Answer 1

The mass of iron(III) hydroxide produced from 24.1 grams of sodium hydroxide is 21.45 grams

Step-by-step explanation:

The equation for the reaction is as follows;

Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(ag)

Molar mass of NaOH = 39.997 g/mol

Molar mass of Fe(OH)₃= 106.867 g/mol

Therefore;4

`Number \ of \ moles \ of \ NaOH = (Mass \ of \ NaOH)/(Molar \ mass \ of \ NaOH)= (24.1)/(39.997 ) = 0.603 \ Moles`

Since 3 moles of NaOH produces 1 mole of Fe(OH)₃

0.603 moles of NaOH will produce 0.603/3 or 0.2008 moles of Fe(OH)₃

∴ Mass of Fe(OH)₃ = Molar mass of Fe(OH)₃ ×Number of moles of Fe(OH)₃

Mass of Fe(OH)₃ = 106.867 g/mol × 0.2008 moles = 21.45 g

The mass of iron(III) hydroxide produced from 24.1 grams of sodium hydroxide = 21.45 grams.

Answer 2

21.4g of Iron III hydroxide

Step-by-step explanation:

The equation of the reaction is;

Fe(NO3)3(aq) + 3NaOH(aq) --------> Fe(OH)3(s) + 3NaNO3(aq)

We know that in solving any problem that has to do with stoichiometry, the balanced reaction equation is indispensable. It serves as a guide in our work.

From the balanced reaction equation;

Number of moles of sodium hydroxide contained in 24.1g of NaOH= mass/molar mass of NaOH

Molar mass of NaOH= 40gmol-1

Number of moles= 24.1/40 = 0.6 moles of NaOH

3 moles of sodium hydroxide produces 1 mole of iron III hydroxide

0.6 moles of sodium hydroxide produces 0.6 ×1/3 = 0.2 moles of iron III hydroxide

Molar mass of iron III hydroxide= 106.867 g/mol

Mass of iron III hydroxide= 0.2 moles × 106.867 g/mol = 21.4g of Iron III hydroxide