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Условие Решение

IOM 2016, Теория — Задача 2. Inorganic transformations and structures

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1st Olympiad of Metropolises — Chemistry, Theoretical Problems, Official English version · 7 септември 2016 г.

Original cover

Country: ______

Name: ______

1st Olympiad of Metropolises

Chemistry

Theoretical Problems

7 September 2016

Moscow, Russia

Chemistry. Theoretical problems. Official English version

Указания

Begin only when the START command is given. You have 4.5 hours to work on the problems.
Use only the pen and calculator provided.
All results must be written in the appropriate boxes within the text. Anything written elsewhere will not be graded. Use the reverse of the problem pages if you need scratch paper.
Write relevant calculations in the appropriate boxes when necessary. If you provide only correct end results for complicated questions, you will receive no score.
Raise your hand if you have any questions concerning the text of the problems.
Raise your hand if you need a restroom break.
The official English version of this examination is available on request only for clarification.

Original key cover

1st Olympiad of Metropolises

Chemistry

Theoretical Problems with solutions

7 September 2016

Moscow, Russia

Chemistry. Theoretical problems with solutions. Official English version

Условие

Problem 2. Inorganic transformations and structures (12 marks)

Question	1	2	3	4	5	6	Total
Points	6	6	2	3	6	1	24

A gas with density $1.50\ \mathrm{g/L}$ (at $273\ \mathrm K$ and $1\ \mathrm{atm}$ ) was passed through a colorless solution containing $1.00\ \mathrm g$ of inorganic salt X that gives a violet flame test. After passage the density of the gas was reduced to $1.43\ \mathrm{g/L}$ (at $273\ \mathrm K$ and $1\ \mathrm{atm}$ ) but the volume remained unchanged. Evaporation of the obtained solution gave $1.289\ \mathrm g$ of a white powder Y that can also be obtained by oxidation of X. If the initial gas is passed through a solution of X in the presence of starch, a violet coloration appears.

1 Identify the initial gas – write the formulas of its components, determine their molar fractions and draw their Lewis structures.

Calculations:

Substances and their molar fractions:

Lewis structures: [6 т.]

2 Determine the formulas of X and Y, write down the reaction equations.

Calculations:

X = ______

Y = ______

Reaction equations: [6 т.]

3 What happens if the water solutions of X and Y are mixed in an acidic medium? Give the ionic equation.

Answer: ______ [2 т.]

4 Draw the geometric structure of Y.

Answer: ______ [3 т.]

5 In an acidic medium from the water solution of Y the white crystals of a salt Z containing 24.6 mass.% of oxygen crystallize. Give a formula of Z and draw the geometric structure of its anion.

Calculations:

Z = ______

Anion structure: [6 т.]

6 What is the use of the initial gas in metropolises?

Answer: ______ [1 т.]

Решение

Official answers and grading scheme

1 Calculations:

Molar mass of the initial gas: $M=\rho V_\mathrm m=1.50\cdot22.4=33.6\ \mathrm{g/mol}$ . There is no individual gas with such molar mass, so we suppose it to be a gas mixture.

Molar mass of the gas after passage through the solution: $M=\rho V_\mathrm m=1.43\cdot22.4=32\ \mathrm{g/mol}$ , this is pure dioxygen, $\mathrm{O_2}$ . The initial mixture consists of ozone $\mathrm{O_3}$ and dioxygen $\mathrm{O_2}$ .

Let $x$ be the molar fraction of $\mathrm{O_2}$ , then

$32x+48(1-x)=33.6$ , $x=0.9$ .

Substances and their molar fractions: $\mathrm{O_2}$ – 90%, $\mathrm{O_3}$ – 10%

Lewis structures:

Each formula – 1 pt,

molar fractions – 2 pts

each Lewis structure – 1 pt

Total – 6 pts

2 Calculations:

The violet flame coloration is typical for potassium salts, a violet coloration of starch indicates that the salt contains iodine. It can be potassium iodide (X). The quantity of this salt should be equal to that of another oxidation product, containing iodine.

So, $n(\mathrm{KI})=1.00/166=0.006024=n(\mathrm Y)$ .

$M(\mathrm Y)=1.289/0{,}006024=214\ \mathrm{g/mol}$ , it corresponds to potassium iodate $\mathrm{KIO_3}$ .

X = $\mathrm{KI}$ Y = $\mathrm{KIO_3}$

Reaction equations:

\mathrm{KI+3O_3=KIO_3+3O_2}

\mathrm{2KI+O_3+H_2O=I_2+O_2+2KOH}

X and Y – 2 pts each (ZERO without calculations)

Reaction equations – 1 pt each (0.5 pts if not balanced)

Total – 6 pts

\mathrm{IO_3^-+5I^-+6H^+=3I_2+3H_2O}

2 pts

4 Draw the geometric structure of Y.

Native iodate geometric structure including lone-pair lobe, bonds, oxygen labels and negative charge.

2 pts

5 Calculation:

The molar mass corresponding to 3 oxygen atoms: $3\cdot16/0.246=195\ \mathrm{g/mol}$ . After subtracting molar mass of $\mathrm{IO_3}$ we get: $195-127-3\cdot16=20\ \mathrm{g/mol}$ . This value is an average between K and H, so Z is an acid iodate $\mathrm{KIO_3\cdot HIO_3}$ , or $\mathrm{KHI_2O_6}$

Z = $\mathrm{KHI_2O_6}$

Anion structure:

Formula of Z – 2 pts (ZERO without calculations)

Anion structure – 2 pts

Total – 4 pts

6 Ozone in mixture with dioxygen is used for water treatment.

1 pt

Бележка от източника

The five native whole theoretical problems and all available official keys are retained. Each original problem heading says 12 marks, while its printed score grid totals 20, 24, 30, 59 or 25; these source contradictions remain visible. In Problem 2 the grid gives 3/6 points for requests 4/5, while the actual key gives 2/4. Problem 3 gives M/N 3/3 points in the answer table and 4/2 in the detailed rubric. Problem 5 request 7 prints 1.5, 1.5 and 2 points but states 4 points in total. Native spelling, formulas, a missing radical dot in one blank answer equation, a missing slash in d[CC]dt, and the missing [M] factor in the printed CC-rate result are preserved without scientific correction. Empty calculation workspaces are represented by their printed response labels and blank fields; molecular answer grids, cyclic-form checkbox tables and scientific axes retain exact original crops. Source question page 10 contains only continuing answer cells; solution pages 18–20 contain the blank repeated Sugars assignment, followed by its detailed key on 21–22.

Оригинал в Архива: Problems.pdf · официални решения: Solutions.pdf