Question

Рассчитайте массу осадка полученного при взаимодействии раствора соли содержащей 15.2 Г сульфата железа (2)FeSo4 с избытком раствора гидроксида натрия NaOH

Answer 1

FeSO4+2NaOH=Fe(OH)2↓+Na2SO4
n(FeSO4)=15.2/152=0.1моль
n(FeSO4)=n(Fe(OH)2)=0.1 моль
m(Fe(OH)3)=0.1*90=9 г

Answer 2

Calculating the Precipitate Mass

To calculate the mass of the precipitate obtained when a solution of iron(II) sulfate containing 15.2 g of FeSO4 reacts with an excess of sodium hydroxide solution (NaOH), we can use the following chemical equation:

FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4

The balanced chemical equation shows that one mole of iron(II) sulfate reacts with two moles of sodium hydroxide to produce one mole of iron(II) hydroxide and one mole of sodium sulfate.

Using the molar masses: - Molar mass of FeSO4 = 151.91 g/mol - Molar mass of NaOH = 40.00 g/mol

We can calculate the mass of the precipitate formed.

The reaction involves a 1:2 ratio of FeSO4 to NaOH. Therefore, the molar amount of NaOH required can be calculated based on the molar amount of FeSO4 available.

The molar amount of FeSO4 can be calculated using the given mass of FeSO4 (15.2 g) and its molar mass.

The molar amount of NaOH can be calculated based on the molar amount of FeSO4 and the stoichiometric ratio in the balanced chemical equation.

Let's calculate the molar amount of FeSO4 and then the molar amount of NaOH.

Molar amount of FeSO4: Molar mass of FeSO4 = 151.91 g/mol Given mass of FeSO4 = 15.2 g

Moles of FeSO4 = Given mass / Molar mass Moles of FeSO4 = 15.2 g / 151.91 g/mol Moles of FeSO4 = 0.1 mol

Molar amount of NaOH: From the balanced chemical equation, 1 mole of FeSO4 reacts with 2 moles of NaOH.

Moles of NaOH = 2 * Moles of FeSO4 Moles of NaOH = 2 * 0.1 mol Moles of NaOH = 0.2 mol

Now, we can calculate the mass of the precipitate formed, which is iron(II) hydroxide (Fe(OH)2).

The molar mass of Fe(OH)2 can be used to calculate the mass of the precipitate formed.

Let's calculate the mass of the precipitate formed.

Mass of Fe(OH)2 (precipitate): Molar mass of Fe(OH)2 = 89.86 g/mol

Mass of Fe(OH)2 = Moles of Fe(OH)2 * Molar mass Mass of Fe(OH)2 = 0.1 mol * 89.86 g/mol Mass of Fe(OH)2 = 8.986 g

Therefore, the mass of the precipitate (iron(II) hydroxide) obtained is 8.986 grams.

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