Answer:
21.4g of Iron III hydroxide
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
Answer:
The mass of iron(III) hydroxide produced from 24.1 grams of sodium hydroxide is 21.45 grams
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
[tex]Number \ of \ moles \ of \ NaOH = \frac{Mass \ of \ NaOH}{Molar \ mass \ of \ NaOH}= \frac{24.1}{39.997 } = 0.603 \ Moles[/tex]
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.
