Solutions of the problems for entrance examinations

Main round

Problem 1.

The required substance is H₂S.

H₂S + 2FeCl₃ = S + 2FeCl₂ + 2HCl

H₂S + 2AgNO₃ = Ag₂S + 2HNO₃

Problem 2.

The methyl radical - CH₃ displays +I-effect and, hence, increases the activity of aromatic system in electrophilic substitution compared to that in benzene. The chlorine atom displays both –I and +M-effects; it reduces the activity of aromatic system, but this reduction is not as strong as in the case of nitro group - NO₂, which displays only negative (–M) electronic effect.

So, the required list is: toluene > chlorobenzene > nitrobenzene.

The reactions are (nitro group - NO₂ directs the substitution into meta-position):

Problem 3.

To solve the problem, we should express the ratio of rate constants at temperatures T₂ and T₁ using the Arrhenius equation. For the first reaction:

Activation energy of the second reaction is half of that of the first reaction: E_A' = E_A/2; hence, the ratio for the second reaction is:

Problem 4.

X - CH₃, Y - COOH.

The reactions:

(5C₆H₅CH₃ + 6KMnO₄ + 9H₂SO₄ = 5C₆H₅COOH + 3K₂SO₄ + 6MnSO₄ + 14H₂O).

Problem 5.

P₂O₃ + 2H₂SO_4(conc) + H₂O = 2H₃PO₄ + 2SO₂ (P⁺³ S⁺⁴)

SO₂ + Ba(OH)₂ = BaSO₃ + H₂O (S⁺⁴ Ba⁺²)

BaSO₃ + 3C = BaS + 3CO (upon heating) (Ba⁺² C⁺²).

Problem 6.

HCOOH + KOH = HCOOK + H₂O

CH₃COOH + KOH = CH₃COOK + H₂O

HCOOK + Ag₂O 2Ag

H₂CO + 2Ag₂O 4Ag

Let (HCOOH) = x, (CH₃COOH) = y, (H₂CO) = z. It follows from reactions with KOH that

(KOH) = x + y = 18.7^. 1.07^. 0.084 / 56 = 0.03 mol.

It follows from reactions with Ag₂O that

(Ag₂O) = 2x + 4z = 9.72 / 108 = 0.09 mol.

The mass of the initial solution is: 46x + 60y + 30z = 2.33 g.

Solving the system of three linear equations, we find: x = 0.005, y = 0.025, z = 0.02.

The molar fractions of the components are: 10% HCOOH, 50% CH₃COOH, 40% H₂CO.

Problem 7.

(Cu) = 4.8 / 64 = 0.075 mol; (NaOH) = 100^. 0.1 / 40 = 0.25 mol.

0.075						0.15
Cu	+	4HNO₃	=	Cu(NO₃)₂	+	2NO₂	+	2H₂O

0.15				0.075		0.075
2NO₂	+	2NaOH	=	NaNO₂	+	NaNO₃	+	H₂O (NaOH is an excess)

The maximum amount of iodine can be obtained through complete reduction of nitrite and nitrate ions to NO:

0.075						0.0375
2NO₂^–	+	2I^–	+	4H⁺	=	I₂	+	2NO	+	2H₂O

0.075						0.1125
2NO₃^–	+	6I^–	+	8H⁺	=	3I₂	+	2NO	+	4H₂O

(I₂) = 0.0375 + 0.1125 = 0.15 mol, m(I₂) = 0.15^. 254 = 38.1 g.