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IChO 1990, Теория — Задача 1. Problem 1 — Phosphoric Acid

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Twenty-second International Chemistry Olympiad — 1990 — Theoretical problems (Volume 2, 2009)

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Attributed archive/compiler teaching source. Retain the Volume 2 attribution and native copying conditions; the printed teaching grant does not establish a broader commercial licence. This 2009 collection is not an unchanged complete jury packet: the editorial preface describes recreated presentation and schemes, numerical corrections, brief keys and omitted practical material. Printed errors, absent measurements, incomplete methods, missing forms and missing answers remain disclosed.

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Anton Sirota (editor), The Competition Problems from the International Chemistry Olympiads, Volume 2, IChO International Information Centre, Bratislava, IUVENTA, 2009.

Native copyright and teaching-copy conditions

THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,

Volume 2
21st – 40th IChO (1989 – 2008)

Editor: Anton Sirota
ISBN 978-80-8072-092-6
Copyright © 2009 by IUVENTA – IChO International Information Centre, Bratislava, Slovakia
You are free to copy, distribute, transmit or adapt this publication or its parts for unlimited teaching purposes, however, you are obliged to attribute your copies, transmissions or adaptations with a reference to "The
Competition Problems from the International Chemistry Olympiads, Volume 2" as it is required in the chemical literature. The above conditions can be waived if you get permission from the copyright holder.

Issued by IUVENTA in 2009
with the financial support of the Ministry of Education of the Slovak Republic

Number of copies: 250
Not for sale.

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Native editorial preface

Preface
Preface
Preface
Preface
This publication contains the competition problems (Volume 2) from the 21st – 40th
International Chemistry Olympiads (IChO) organized in the years 1989 – 2008 and is a continuation of the publication that appeared last year as Volume 1 and contained competition problems from the first twenty IChOs. The whole review of the competition tasks set in the IChO in its fourty-year history is a contribution of the IChO International
Information Centre in Bratislava (Slovakia) to the development of this world known international competition. This Volume 2 contains 154 theoretical and 46 practical competition problems from the mentioned years. The review as a whole presents altogether 279 theoretical and 96 practical problems.
In the elaboration of this collection the editor had to face certain difficulties because the aim was not only to make use of past recordings but also to give them such a form that they may be used in practice and further chemical education. Consequently, it was necessary to make some corrections in order to unify the form of the problems. However, they did not concern the contents and language of the problems.
Unfortunately, the authors of the particular competition problems are not known and due to the procedure of the creation of the IChO competition problems, it is impossible to assign any author's name to a particular problem. As the editor I would appreciate many
times some discussion with the authors about any critical places that occurred in the text.
On the other hand, any additional amendments to the text would be not correct from the
historical point of view. Therefore, responsibility for the scientific content and language of the problems lies exclusively with the organizers of the particular International Chemistry
Olympiads.

Some parts of texts, especially those gained as scanned materials, could not be used directly and thus, several texts, schemes and pictures had to be re-written or created again. Some solutions were often available in a brief form and necessary extent only, just for the needs of members of the International Jury.

Recalculations of the solutions were made in some special cases only when the numeric results in the original solutions showed to be obviously not correct. Although the numbers of significant figures in the results of several solutions do not obey the criteria generally accepted, they were left without change.

In this publication SI quantities and units are used and a more modern method of chemical calculations is introduced. Only some exceptions have been made when, in an effort to preserve the original text, the quantities and units have been used that are not SI.
There were some problems with the presentation of the solutions of practical tasks, because most of the relatively simple calculations were based on the experimental results of contestants. Moreover, the practical problems are accompanied with answer sheets in the last years and several additional questions and tasks have appeared in them that were not a part of the text of the original experimental problems. Naturally, answer sheets could not be included in this publication and can only be preserved as archive materials.

When reading the texts of the IChO problems one must admire and appreciate the work of those many known and unknown people – teachers, authors, pupils, and organizers – who contributed so much to development and success of this important international competition.

I am sure about the usefulness of the this review of the IChO problems. It may serve not only as archive material but, in particular, this review should serve to both competitors and their teachers as a source of further inspiration in their preparation for this challenging competition.

Bratislava, July 2009

Anton Sirota, editor

Условие

The elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\mathrm{Ca^{2+}}$ , $\mathrm{CO_3^{2-}}$ , $\mathrm{SO_4^{2-}}$ , $\mathrm{SiO_3^{2-}}$ , and $\mathrm{F^-}$ .

Let us assume that this mineral is a mixture of tricalcium phosphate, $\mathrm{Ca_3(PO_4)_2}$ , calcium sulphate, calcium fluoride, calcium carbonate and silica.

For uses as fertilizer the calcium bis(dihydrogenphosphate), $\mathrm{Ca(H_2PO_4)_2}$ , which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.

The elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:

Component	$\mathrm{CaO}$	$\mathrm{P_2O_5}$	$\mathrm{SiO_2}$	F	$\mathrm{SO_3}$	$\mathrm{CO_2}$
% by mass	47.3	28.4	3.4	3.4	3.5	6.1

Operation 1. A sample of $m_0$ of this mineral is treated with $50.0\ \mathrm{cm^3}$ of a solution containing $0.500\ \mathrm{mol\,dm^{-3}}$ phosphoric and $0.100\ \mathrm{mol\,dm^{-3}}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70\,{}^\circ\mathrm{C}$ avoiding temperature rising above $90\,{}^\circ\mathrm{C}$ . This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_1$ is the mass of the residue obtained. In these conditions only dihydrogenphosphate, $\mathrm{Ca(H_2PO_4)_2}$ , is formed while silica and silicate do not react.

Operation 2. $1.00\ \mathrm{g}$ of this residue is treated with $50.0\ \mathrm{cm^3}$ of water at $40\,{}^\circ\mathrm{C}$ , then filtered, dried and weighed. The mass of the residue obtained is $m_2$ . This new residue is mainly containing gypsum, $\mathrm{CaSO_4\cdot2H_2O}$ , whose solubility can be considered as constant between $20\,{}^\circ\mathrm{C}$ and $50\,{}^\circ\mathrm{C}$ and is equal to $2.3\ \mathrm{g\,dm^{-3}}$ .

Questions

1.1
Write the balanced equations for the reactions that are involved.

1.2
From what mass of apatite should one start if all the reactions are stoichiometric?

Starting with $m_0$ of obtained apatite, $m_1=5.49\ \mathrm{g}$ of residue are obtained.

1.3
What mass should theoretically be obtained?

1.4
This result is due to the presence of products that are not expected to be found in the residue. Give two of them that under these experimental conditions can plausibly account for the data.

Traditionally, in industry the analysis and the yield are expressed as percentage of oxide. The phosphorous content is expressed as if it were $\mathrm{P_2O_5}$ .

If $n_2$ is the amount of a soluble product obtained, $n_1$ the amount of a substance added as acid, $n_0$ the amount of apatite added, the yield is:

r_{\mathrm{exp}}=\frac{n_2}{n_1+n_0}\,100.

$m_2=0.144\ \mathrm{g}$ of residue is obtained on the filter.

1.5
Calculate $r_{\mathrm{exp}}$ .

1.6
The experimental yield is over 100%. Calculate a value of $r$ nearer to the real yield.

Relative atomic masses: P: 31; Ca: 40; O: 16; H: 1; F: 19; C: 12; Si: 28; S: 32.

Values of $pK$ :

Acid/base pair	$pK$
$\mathrm{HSO_4^-}/\mathrm{SO_4^{2-}}$	2
$\mathrm{HF}/\mathrm{F^-}$	3
$\mathrm{H_3PO_4}/\mathrm{H_2PO_4^-}$	2
$\mathrm{H_2PO_4^-}/\mathrm{HPO_4^{2-}}$	7
$\mathrm{HPO_4^{2-}}/\mathrm{PO_4^{3-}}$	12

Source notes

The first printed reaction produces HF with no F on its left. The third prints water separately on both sides. The $\mathrm{CaF_2}$ denominator $2+19$ , factors $1.99/1.9$ , products $1.80\ \mathrm{mol}$ and $2.77\ \mathrm{mol}$ , dimensional equality for $m$ , $n_2$ exponent, $0.741\ \mathrm{mol}$ and native placement of 96% under 1.5 versus 100% under 1.6 are all retained without repair. The residual $\mathrm{CaO}$ exponent is clearly negative four; the earlier reader-note allegation of positive four was superseded by native-image zoom.

Решение

1.1

The following equations are transcribed literally from the collection:

\begin{aligned} \mathrm{Ca_3(PO_4)_2}+4\mathrm{H_3PO_4}&\longrightarrow\mathrm{Ca(H_2PO_4)_2}+2\mathrm{HF},\\ \mathrm{CaCO_3}+2\mathrm{H_3PO_4}&\longrightarrow\mathrm{Ca(H_2PO_4)_2}+\mathrm{CO_2}\uparrow+\mathrm{H_2O},\\ \mathrm{Ca_3(PO_4)_2}+2\mathrm{H_2SO_4}+4\mathrm{H_2O}&\longrightarrow2\mathrm{CaSO_4}+2\mathrm{H_2O}+\mathrm{Ca(H_2PO_4)_2},\\ \mathrm{CaF_2}+\mathrm{H_2SO_4}+2\mathrm{H_2O}&\longrightarrow\mathrm{CaSO_4\cdot2H_2O}+2\mathrm{HF}\uparrow,\\ \mathrm{CaCO_3}+\mathrm{H_2SO_4}+\mathrm{H_2O}&\longrightarrow\mathrm{CaSO_4\cdot2H_2O}+\mathrm{CO_2}\uparrow. \end{aligned}

1.2

$1\ \mathrm{g}$ of apatite contains:

\frac{0.284}{142}=2.00\times10^{-3}\ \mathrm{mol\ of\ Ca_3(PO_4)_2},

\frac{0.034}{2+19}=0.89\times10^{-3}\ \mathrm{mol\ of\ CaF_2}, \qquad \frac{0.061}{44}=1.39\times10^{-3}\ \mathrm{mol\ of\ CaCO_3},

\frac{0.035}{80}=0.44\times10^{-3}\ \mathrm{mol\ of\ CaSO_4},

\frac{0.473}{56}-\frac{3\times0.284}{142} -\frac{0.034}{38}-\frac{0.061}{44}-\frac{0.035}{80} =2.72\times10^{-4}\ \mathrm{mol\ CaO\ that\ remain}.

The amount of $\mathrm{H_3PO_4}$ needed to react with $1\ \mathrm{g}$ of apatite is equal to:

n(\mathrm{H_3PO_4}) =4n(\mathrm{Ca_3(PO_4)_2})+2n(\mathrm{CaF_2})+2n(\mathrm{CaCO_3}) =12.56\times10^{-3}\ \mathrm{mol}.

$50\ \mathrm{cm^3}$ of the acid contains $25\times10^{-3}\ \mathrm{mol}$ of $\mathrm{H_3PO_4}$ , therefore $25/12.56=1.99\ \mathrm{g}$ apatite is needed to neutralize the $\mathrm{H_3PO_4}$ present.

The amount of $\mathrm{H_2SO_4}$ needed to react with $1\ \mathrm{g}$ of apatite can be calculated in the same way:

n(\mathrm{H_2SO_4}) =2n(\mathrm{Ca_3(PO_4)_2})+n(\mathrm{CaF_2})+n(\mathrm{CaCO_3}) =6.28\times10^{-3}\ \mathrm{mol}.

$50\ \mathrm{cm^3}$ of the acid contains $5.00\times10^{-3}\ \mathrm{mol}$ of sulphuric acid. Therefore $5/6.28=0.80\ \mathrm{g}$ of apatite is needed to neutralize the $\mathrm{H_2SO_4}$ . The total amount of apatite is $m_0=1.99+0.80=2.79\ \mathrm{g}$ .

1.3

Formation of $\mathrm{Ca(H_2PO_4)_2}$ :

$1.99\ \mathrm{g}$ of apatite needed to neutralize the $\mathrm{H_3PO_4}$ contains $1.9\times2.00\times10^{-3}\ \mathrm{mol}$ of $\mathrm{Ca_3(PO_4)_2}$ , thus $3\times2\times2\times10^{-3}=1.2\times10^{-2}\ \mathrm{mol}$ of dihydrogen phosphate is being formed.

From $\mathrm{CaF_2}$ , $1.99\times0.89=1.80\ \mathrm{mol}$ and from $\mathrm{CaCO_3}$ , $1.99\times1.39=2.77\ \mathrm{mol}$ of $\mathrm{Ca(H_2PO_4)_2}$ are formed.

$0.8\ \mathrm{g}$ of apatite that reacts with $50\ \mathrm{cm^3}$ of the sulphuric acid yields $2\times0.8\times10^{-3}=1.6\times10^{-3}\ \mathrm{mol}$ of $\mathrm{Ca(H_2PO_4)_2}$ .

The printed dimensional equality is:

m(\mathrm{Ca(H_2PO_4)_2})=18.07\times10^{-3}\ \mathrm{mol}=4.230\ \mathrm{g}.

Formation of gypsum:

n(\mathrm{CaSO_4})=n(\mathrm{H_2SO_4})=5.00\times10^{-3}\ \mathrm{mol} \ \widehat{=}\ 0.86\ \mathrm{g}.

The amount of $\mathrm{CaSO_4}$ that was already present in $1\ \mathrm{g}$ of apatite and yielded gypsum is $0.434\times10^{-3}\times172=0.075\ \mathrm{g}$ . There remain also $0.034\ \mathrm{g}$ of silica, and thus the theoretical mass of the residue should be:

m_{\mathrm{th}}=4.230+0.86+(0.0753+0.034)\times2.79=5.39\ \mathrm{g}.

1.4

The difference of $0.1\ \mathrm{g}$ may be due to water and unreacted $\mathrm{CaF_2}$ in the residue.

1.5

The second reaction is intended to dissolve $\mathrm{Ca(H_2PO_4)_2}$ , while all the other products remain on the filter.

According to the yielded residue of $0.144\ \mathrm{g}$ , $1\ \mathrm{g}$ of residue contains $1-0.144=0.856\ \mathrm{g}$ of soluble product. If it were all $\mathrm{Ca(H_2PO_4)_2}$ it would correspond to $0.856/234=3.66\times10^{-3}\ \mathrm{mol}$ . For $5.49\ \mathrm{g}$ of residue it is $0.0201\times10^{-3}\ \mathrm{mol}$ of soluble product ( $n_2$ ). The amount of acid used is $0.500/20=0.025\ \mathrm{mol}\ \mathrm{H_3PO_4}$ (equals $0.0125\ \mathrm{mol}\ \mathrm{P_2O_5}$ ) and $0.005\ \mathrm{mol}\ \mathrm{H_2SO_4}$ . The amount of $\mathrm{Ca_3(PO_4)_2}$ in $2.79\ \mathrm{g}$ apatite is $0.00558\ \mathrm{mol}$ (equals $0.00558\ \mathrm{mol}\ \mathrm{P_2O_5}$ ). So:

r_{\mathrm{exp}}=100\times\frac{0.0201}{0.0125+0.00558}=111\%.

Since $50\ \mathrm{cm^3}$ water dissolve $0.115\ \mathrm{g}$ of gypsum, the real quantity of $\mathrm{Ca(H_2PO_4)_2}$ is $0.856-0.115=0.741\ \mathrm{mol}$ , so that the real yield gives:

r_{\mathrm{exp}}=100\times\frac{0.0174}{0.0125+0.00558}=96\%.

1.6

The theoretical value for $r_{\mathrm{exp}}$ is:

r_{\mathrm{exp}} =100\times\frac{4.23/234}{0.0125+0.00558}=100\%,

so this calculation makes sense.

Оригинал в Архива: IChO 21st-40th.pdf · решение от архива: IChO 21st-40th.pdf · допълнителни данни: IChO 21st-40th.pdf · допълнителни данни: IChO 21st-40th.pdf