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2 years ago

KFell Fe"(CN), + e + Nat → KNaFe'Fe(CN)6

Chemistry

1 answer:

Alinara [238K]2 years ago

7 0

Answer:

Most common oxidation states: +2, +3

M.P. 1535º

B.P. 2750º

Density 7.87 g/cm3

Characteristics: Iron is a gray, moderately active metal.

Characteristic reactions of Fe²⁺ and Fe³⁺

The [Fe(H2O)6]3+ ion is colorless (or pale pink), but many solutions containing this ion are yellow or amber-colored because of hydrolysis. Iron in both oxidation states forms many complex ions.

Aqueous Ammonia

Aqueous ammonia reacts with Fe(II) ions to produce white gelatinous Fe(OH)2, which oxidizes to form red-brown Fe(OH)3:

Fe2+(aq)+2NH3(aq)+3H2O(l)↽−−⇀Fe(OH)2(s)+2NH+4(aq)(1)

Fe3appt.gif

Aqueous ammonia reacts with Fe(III) ions to produce red-brown Fe(OH)3:

Fe3+(aq)+3NH3(aq)+3H2O(l)↽−−⇀Fe(OH)3(s)+3NH+4(aq)(2)

Fe3bppt.gif

Both precipitates are insoluble in excess aqueous ammonia. Iron(II) hydroxide quickly oxidizes to Fe(OH)3 in the presence of air or other oxidizing agents.

Sodium Hydroxide

Sodium hydroxide also produces Fe(OH)2 and Fe(OH)3 from the corresponding oxidation states of iron in aqueous solution.

Fe2+(aq)+2OH−(aq)↽−−⇀Fe(OH)2(s)(3)

Fe4appt.gif

Fe3+(aq)+3OH−(aq)↽−−⇀Fe(OH)3(s)(4)

Fe4bppt.gif

Neither hydroxide precipitate dissolves in excess sodium hydroxide.

Potassium Ferrocyanide

Potassium ferrocyanide will react with Fe3+ solution to produce a dark blue precipitate called Prussian blue:

K+(aq)+Fe3+(aq)+[Fe(CN)6]4−(aq)↽−−⇀KFe[Fe(CN)6](s)(5)

Fe5a1ppt.gif

With Fe2+ solution, a white precipitate will be formed that will be converted to blue due to the oxidation by oxygen in air:

2Fe2+(aq)+[Fe(CN)6]4−(aq)↽−−⇀Fe2[Fe(CN)6](s)(6)

Fe5a2ppt.gif

Many metal ions form ferrocyanide precipitates, so potassium ferrocyanide is not a good reagent for separating metal ions. It is used more commonly as a confirmatory test.

Potassium Ferricyanide

Potassium ferricyanide will give a brown coloration but no precipitate with Fe3+. With Fe2+, a dark blue precipitate is formed. Although this precipitate is known as Turnbull's blue, it is identical with Prussian blue (from Equation 5).

K+(aq)+Fe+2(aq)+[Fe(CN)6]3−(aq)↽−−⇀KFe[Fe(CN)6](s)(7)

Fe5b.gif

Potassium Thiocyanate

KSCN will give a deep red coloration to solutions containing Fe3+:

Fe+3(aq)+NCS−(aq)↽−−⇀[FeNCS]+2(aq)(8)

Fe5cppt.gif

No Reaction

Cl−, SO2−4

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2 years ago

Rank the following fertilizers in decreasing order of mass percentage of nitrogen:

charle [14.2K]

<h3>Answer:</h3>

NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄ = 132.06 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age = 21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age = 21.18 %

In KNO₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of KNO₃ = 101.10 g.mol⁻¹

Mass %age = Mass of N / M.Mass of KNO₃ × 100

Mass %age = 14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age = 13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄ = 115.03 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age = 14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age = 12.17 %

In NH₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of NH₃ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₃ × 100

Mass %age = 14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age = 82.20 %

In NH₄NO₃:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of NH₄NO₃ = 80.04 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age = 28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age = 34.98 %

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