OLIMPIADI DELLA CHIMICA (IN INGLESE)

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THE FIRST

INTERNATIONAL CHEMISTRY OLYMPIAD (Praga 1968)

THEORETICAL PROBLEMS

PROBLEM 1

A mixture of hydrogen and chlorine kept in a closed flask at a constant temperature was irradiated by scattered light. After a certain time the chlorine content decreased by 20 % compared with that of the starting mixture and the resulting mixture had the composition as follows: 60 volume % of chlrine, 10 volume % of hydrogen, and 30 volume % of hydrogen chloride.

Problems:

1.1 What is the composition of the initial gaseous mixture?

1.2 How chlorine, hydrogen, and hydrogen chloride are produced?

 

SOLUTION

1.1 The chemical reaction is 

H2+ Cl2—-> 2 HCl

From this reaction  you can see that  if 30 % of Hydrogen Chloride was formed it  could only be formed by the reaction of 15 volume parts of hydrogen and 15 volume parts of chlorine.

If Hydrogen at end of reaction is 10% and 15% went to reaction then initial quantity of H was 10+15 =25%

If Chlorin at end of reaction is 60%  and 15 % went to form HCl  then initial quantity was 60 +15 = 75%

Hence, the initial composition of the mixture had to be:

Cl,: 60 + 15 = 75 %

H2: 10+15=25%

 

1.2 Chlorine and hydrogen are produced by electrolysis of aqueous solutions of NaCI:       

NaCl(aq) —-> Na +(aq) + Cl (aq)

anode: (positive side)

2 Cl – 2 e —-> Cl2

cathode:(negative side)

2 Na + + 2 e —-> 2 Na

but at cathod the Na formed, go to reaction with H2O and there is another  reaction:

2 Na + 2 H2O —> 2 NaOH + H2

H2 and Cl2 forms 2HCl

PROBLEM 2

Write down equations for the following reactions:

2.1 Oxidation of chromium(III) chloride with bromine in alkaline solution (KOH).

2.2 Oxidation of potassium nitrite with potassium permanganate in acid solulion (H2S04).

2.3 Action of chlorine on lime water (Ca(OH)2) in a cold reaction mixture.

SOLUTION

2.1

the first  is a redox reaction where Cr(III) becoms  Cr(VI)  (oxidation semi reaction) and Br2 (n.o.zero) becoms Br- (n.o. -1)

2[Cr(III)  + 8 OH-       →  CrO4-2 + 4H2O + 3 e

3[Br2  +2 e                    → 2 Br-

________________________________

2 Cr(III) + 3 Br2 + 16 OH-  →  2 CrO4-2 + 6r- +8H2O

now we can write the reaction

2 CrCI3 + 3 Br2+ 16 KOH ➔ 2 K2Cr04 + 6 KBr + 6 KCI + 8 H2O

2.2

this is also a redox reaction  where N (III) becoms  N(V)  (oxidation) and Mn(VII) becoms Mn(II) (reduction) with H+ (from H2SO4)

 

5[NO2-   + H2O             NO3-  +2H+  + 2 e

2[MnO4-  + 8 H+ + 5e      Mn+2  + 4 H2O

_____________________________________

5NO2- + 2MnO4- + 5 H2O + 16 H+ ➔   5NO3- + 2 Mn+2 + 10H+ + 8 H2O

simplifying   we have

5NO2- + 2MnO4-   + 6 H+ ➔   5NO3- + 2 Mn+2  + 3 H2O

Now we can write the reaction

( remeber H+ comes from H2SO4 then 6 H+ comes from 3H2SO4 and SO4= react with K+ andMn+2)

5 KNO2 + 2 KMnO4 + 3 H2SO4 ➔ 2 MnSO4 + K2SO4 + 5 KNO3 + 3 H2O

2.3.

The last reaction is also redox reaction called dismutation because  Cl2  in a cold reaction mixture forms Cl-1 and Cl+1.

In effect CaOCl2 can be writed  Ca(OCl)Cl

Here one of chlorine atom is connected to calcium and one to oxygen atom.

Note : It is not calcium hypochlorite but calcium hypochlorite chloride. Only Ca(OCl)2 is called  calcium hypochlorite.

At last we can write the reaction

 Cl2+ Ca(OH)2➔ CaOCl2 + H2O   but in a cold reaction mixture

PROBLEM 3 (difficile) 

The gas escaping from a blast furnace has the following composition:

12.0 volume % of CO2     28.0 volume % of CO

3.0 volume % of H2            0.6 volume % of CH4

0.2 volume% of C2H4    56.2 volume % of N2

3.1    Calculate the theoretical consumption of air (in m3) which is necessary for a total combustion of 200 m3 of the above gas if both the gas and air are measured at the same temperature. (Oxygen conteni in the air is about 20 % by volume).

3.2    Determina the composition of combustion products if the gas is burned in a 20 % excess of air.

solution

the reaction 2CO + O2→ 2CO2        shows that we need 1 part of O2 for 2 parts of Co then for 28 parts of CO we need 14 of O2

next reaction 2H2 +O2 → 2H2O     shows that we need 1 part of O2 for 2 parts of H2  then for 3 parts of H2 we need   1,5 of O2

next reaction CH4 +2O2 → CO2 +2H2O  shows that we need 2 part of O2 for 1 parts of CH4  then for 0,2 parts of CH4 we need   0,6 of O2

next reaction C2H4 +3O2 → 2CO2 + 2 H2O shows that we need 3 part of O2 for 1 parts of C2H4  then for 0,6 parts of CH4 we need   1,2 of O2

we have to remember that O2  is 20% in the air  not 100%  for any reaction then we need  14 + 1,5 + 1,2+ 0,6 = 17,3 parts x 5 = 86.5 part of air ( x5 is because 20 % x 5 =100 %).

this amount of O2 is for 100 m3  of mixture then for 200 m3   we need 2 x 86,5 parts = 173 m3 of air

but  we have to add  oxygen then 20% of 173 m3  = 34,6 m3

Finally we need 173 + 34,6 = 207,6 m3 of air 

3.2

we know that 207,6 represents 5 times 20% of air and that 20% that is 2 parts compared to 5 which is the total is 0.2 x 206.7 = 41.52 
therefore

206,7 :5=41,52 parts of O2

41,52 /2 = 20,76 parts of O2 for 100 m3 of the gas   

20.76 x 4 = 83.04 parts of N2 for 100 m3 of the gas

to give answere to the second  part of the problem we see that CO2 is 12 % volume

2CO + O2→ 2CO2   the volume parts are 2 times of O2 then 2 x 14 =28

CH4 +2O2 → CO2 +2H2O  the volume parts are 1/2 of O2 then  1/2 •1,2 =0,6

C2H4 +3O2 → 2CO2 + 2 H2O  the volume parts are  2/3 of O2 then 2/3• 0,6 = 0,4

Total volume parts of CO2= 12+28+0,6+0,4 = 41

for H2O we have 3 +1,2+0,4 =4,6 

for N2 we have 56,2 + 83,04= 139,24

For O2  we have  20,76 -17,30= 3,46 

total volume is 139,24 +4,6+41+3,46 = 188,3 

207.6 : 5 = 41.52 parts of 02

41,52 : 2 = 20. 76 parts of 0for 100 m3 of the gas

The parts of N2 for 100 m3 of the gas are 80% the 4 times 20% then

parts N2 are 20.76 x 4 = 83.04 

finally we can calculate the composition % of the mixture

% O2 = 3,46/188,3 x 100= 1,84 % 

 % N2 = 139,24 /188,3 x 100 = 73,95% 

  % H2O = 4,6/188,3 x 100 = 2,44 %

%CO2 = 41/ 188,3 x 100  = 21,77 %

PROBLEM4

A volume of 31.7 cm3 of a 0.1-normal NaOH is required lor the neutralization of 0.19 g of an organic acid whose vapour is thirty times as dense as gaseous hyrogen. Problem:

4.1 Give the name and structural formula of the acid. (The acid concerned is a common organic acid.)

 

SOLUTION 4.1

a) We know that volume(ml)  x Molarity = millimoles  or volume (L) x Molarity = Moles

then moles of NaOH = 0.1 mol /L x 0.0317 L = 3.17 x 103mol

we know also that volume1 x M1 = volume2 x M2 then

moles of acid are m= 3.17 x 103mol

But grams /MW = moles then  0,19 / 3.17 x 103mol =  Acid Molar Weight =60

b)   From the ideal gas law we know that density is proportional to Molecular weight then d= K M  k= d/M 

d1/M1 = d2/M2

d1/d2 = M1/M2

but   M1 = 2 g/mol and d1 = 1

M2 =X  d2= 30 x 1

1/30 = 2 /X      X= 30 x 2 =60

The acid concerned is a monoprotic acid and its molar mass is 60 g /mol.

The acid is acetic acid: CH3-COOH

THE SECOND

INTERNATIONAL CHEMISTRY OLYMPIAD ( Katowice 1969)

THEORETICAL PROBLEMS

PROBLEM 1

An amount of 20 g of potassium sulphate was dissolved in 150 cm3 of water. The solution was then electrolysed. Alter electrolysis, the content of potassium sulphate in the solution was 15 % by mass.

Problem:

What volumes of hydrogen and oxygen were obtained at a temperature of 20 °C and a pressure of 101 325 Pa?

SOLUTION

On electrolysis, only water is decomposed and the total amount of potassium sulphate in the  solution is constant.

The mass of water in the solution:

1.1 Before electrolysis (on the assumption that  d = 1 g / cc): m(H2O) = 150 g

1.2 After electrolysis:

the  solution mass is 20/0,15 = 133,3

m(H2O)=m(solution)-m(K2SO4) = 150-133,3 = 36,7

Then the mass of water decomposed on electrolysis: m(H2O) = 150 -113.3 = 36.7 g, i. e.

but Grams/MW = moles

moles(H2O)  = 2.04 mol

Since, 2 H2O →2 H2 + O2  thus,

n(H2) = 2.04 mol

n(O2) = 1.02 mal

from the gas law VH2 = 2,04 x8,314  J mol-1 x 293,15 / 101325  = 0,049 m3 or 49 L

Vo2  = 1/2 VH2 = 24,5 L

PROBLEM 2
A compound A contains 38.67 % of potassium, 13.85 % of nitrogen, and 47.48 % of oxygen. On heating, it is converted to a compound B containing 45.85 % of potassium, 16.47 % of nitrogen, and 37.66 % of oxygen.
Problem:
2.1 What are the stoichiometric formulas of the compounds?
2.2 Write the corresponding chemical equation

solution

to calculate the compound’s formula we have to divide the %  by the respective atomic weight and we obtain :

38.67 %/ 39,1  =  0,989  K    13.85 %/14  =0,989 N    47.48 %/ 16 = 2,968  O

we divide each result by the smallest number obtained that is 0,989 and we have the formula   K1 N1 O3  or KNO3
after heating the % composition is  changed  then there is another compound and we repeat the same calculation
 45.85 % /39,1 = 1,173   K    16.47 %/14  = 1,176 N  and 37.66 % of oxygen.
we divide each result by the smallest number obtained that is 1,173 and we have the formula K1N1O2 or KNO2
Finally the reaction after heating is     2KNO3 → 2KNO2 + O2 ↑

PROBLEM 3
A 10 cm3 sample of an unknown gaseous hydrocarbon was mixed with 70 cm3 of oxygen and the mixture was set on fire by means of an electric spark. When the reaction was over and water vapours were liquefied, the final volume of gases decreased to 65 cm3. This mixture

then reacted with a potassium hydroxide solution and the volume of gases decreased to 45 cm3
.
Problem:
What is the molecular formula of the unknown hydrocarbon if volumes of gases were measured at standard temperature and pressure (STP) conditions?

solution

The unknown gaseous hydrocarbon has the general formula: CxHy

Now we have to remember that at standard condition  1 mole of any gas have a 22,414 L  then  X moles of sample  are  in 10 cm3

VCxHy  = 1x 0,01 / 22,414 = 4,46 x 10-4 moles CxHy

Balance of oxygen:
– Before the reaction:  0.070/22.4 mol
– After the reaction: 45 cm3  0.045/22.4 mol
Consumed in the reaction: 0.025/22.4 mol of O2
According to the equation:
CxHy + (x + y/4) O2 = x CO2 +y/2 H2O
Hence, 0.020/22.4 mol of O2 reacted with carbon and 0.020/22.4 mol of CO2 was formed

(C + O2 = CO2),

0.005/22.4 mol O2 combined with hydrogen and 0.010/ 22.4 mol of water was obtained

(2 H2 + O2 = 2 H2O).

3 n(C) = n(CO2) = 0.020/22.4 mol
n(H2) = 2 n(H2O) = 0.020/22.4 mol
x : y = n(C) : n(H2) = 0.020 : 0.020 = 1 : 1
From the possible solutions C2H2, C3H3, C4H4, C5H5.only C2H2 satisfies to the conditions given in the task, i. e. the unknown hydrocarbon is acetylene CH≡CH

PROBLEM 4
Calcium carbide and water are the basic raw materials in the production of:
a) ethanol
b) acetic acid
c) ethylene and polyethylene
d) vinyl chloride
e) benzene
Problem:
Give basic chemical equations for each reaction by which the above mentioned compounds can be obtained.

solution

Basic reaction: CaC2 + 2 H2O = Ca(OH)2 + C2H2
From acetylene can be obtained:
a) ethanol
CH ≡CH   + H2O + HgSO4 (catalyst) diluted H2SO4     —–→  CH2= CH-OH (alcol vinilico) —-> CH3CHO (acetaldeide) —reduction-> CH3CH2OH

b) acetic acid

CH ≡CH   + H2O + HgSO4 (catalyst) diluted H2SO4     —–→  CH2= CH-OH (alcol vinilico) —-> CH3CHO (acetaldeide) —oxydation->

CH3COOH

c) ethylene, polyethylene

CH≡ CH + H2O   + catalyst—>CH2=CH2 (ethylene)  + pressure and catalyst——> -(CH-CH-)-   polyethylene

d) vinyl chloride
CH≡ CH   + HCl—–> CH2= CH- Cl  vinyl chloride

e) benzene

3 CH≡ CH  400 -500 °C —->  C6H6 benzene     

 

THE THIRD

INTERNATIONAL CHEMISTRY OLYMPIAD ( Budapest, 1970) 

THEORETICAL PROBLEMS
PROBLEM 1
An amount of 23 g of gas (density ρ = 2.05 g dm-3 at STP) when burned, gives 44 g of carbon dioxide and 27 g of water.
Problem:
What is the structural formula of the gas (compound)?

SOLUT ION
We call  X the unknown gas.

We can calculate the Molecular weight M(X)  from the ideal gas law : M(X) = d(x) R T / p   = 46 g/ mol

Now we calculate the moles of Gas X

n (X)=  23 /46 g mol-1 = 0,5  mol

we also can calculate the moles of CO2 and

n(CO )2 = 44 g/44 g mol-1  =  1 mol

The we know that   1 mole of CO2 (44 g)  contain  1 C  (12 g)  and then
n(C) = 1 mol
m(C) = 12 g

The same calculation we do for H2O

n(H2 O) =27 g / 18 g mol-1   = 1,5 mol
n(H) = 3 mol
m(H) = 3 g

Because the compound contains also oxygen,  His quantity can be calculated

grams  of compounds = grams of C + grams of H + grams of O    then

23 g  =  12 g C+ 3 g H + x g O

g (O) = 23 g – 15 g = 8 g O   then moles of O are  8/16 = 0,5

n(O) = 0,5 mol

the empirical formula is :

n(C) : n(H) : n(O)

1 : 3 : 0,5      then      2 : 6 : 1
The empirical formula of the compound is C2H6O.

C2H6O—–> C2H5OH ethanol

C2H6O —-> CH3-O- CH3   dimethyl ether

to decide if is ethanol  or dimethyl ether we know that ethanol is liquid  then the compound is  dimethyl ether

PROBLEM 2
A sample of crystalline soda (A) with a mass of 1.287 g was allowed to react with an excess of hydrochloric acid and 100.8 cm3 of a gas was liberated (measured at STP). Another sample of different crystalline soda (B) with a mass of 0.715 g was decomposed by 50 cm3 of 0.2 N sulphuric acid. After total decomposition of soda, the excess of the sulphuric acid was neutralized which required 50 cm3 of 0.1 N sodium hydroxide solution (by titration on methyl orange indicator).

Problems

2.1 How many molecules of water in relation to one molecule of Na2CO3 are contained in the first sample of soda?
2.2 Have both samples of soda the same composition? Relative atomic masses: Ar(Na) = 23; Ar(H) = 1; Ar(C) = 12; Ar(O) = 16.

solution

2.1  Sample A

we know that at STP condition, for  1 mole of gas we have  22,414 liters  then  for 0,1008 L  we have 0,0045 mol

but Na2CO3 . x H2O    m(A) = 1.287 g then 

n(CO2)= 0,0045 mol   the the Molecular weight is  g/mol = MW

M(A) = 1,287 / 0,0045 = 286 g /mol

We know that

M(A) = M(Na2CO3) + x M(H2O)

then to calculate x 

X=M(A) –  M(Na2CO3)  / M(H2O)

X= (286- 106) g mol-1 / 18 g mol-1 = 10

the sample A is  : Na2CO3 .10 H2O

2.2 Sample B: Na2CO3 . x H2O

m(B) = 0.715 g

to solve the first problem we calculate the excess of H2SO4  remembering that

volume(ml)  x N = mequivalents   and for H2SO4  mol= equiv/ 2    (there are 2 H atoms in H2SO4)

50 x 0,1 = 5 meq

n(NaOH)  =50 x 0,1 = 5 meq = 0.005 mol

and for H2SO4  mol= equiv/ 2    (there are 2 H atoms in H2SO4)

Excess of H2SO4: n(H2SO4) = 0.0025 mol

The amount of substance combined with sample B:
n(H2SO4) = 0.0025 mol = n(B)

M(B) = 0,715 g/ 0.0025 g mol-1= 286 g mol-1

Sample B: Na2CO3 .10 H2O

PROBLEM 3

Carbon monoxide was mixed with 1.5 times greater volume of water vapours. What will be the composition (in mass as well as in volume %) of the gaseous mixture in the equilibrium state if 80 % of carbon monoxide is converted to carbon dioxide?

solution

the reaction is :
CO + H2O→ CO2 + H2

to calculate % Volume and 
because the problem do not give us any quantity we can suppose :
n(CO) = 1 mol
n(H2O) = 1.5 mol

then  after reaction:
n(CO) = 0.2 mol  remaining because 80% reacted
n(H2O) = 0.7 mol remaining (1,5 mol – 0,8 mol)
n(CO2) = 0.8 mol formed
n(H2) = 0.8 mol formed

then  if  0,2 mol of CO remain   from a total quantity of 1+1,5 mol =2,5 mol the % is :

2,5 : 0,2 = 100 :X           X= 20/2,5 = 8%  CO

if 0,7 mol H2O remain from a total quantity of 1+1,5 mol =2,5 mol the % is

2,5 : 0,7 = 100 : X          X= 28 %  H2O

and   0,8 x 100/ 2,5 =   32% CO2

also  0,8 x 100 /2,5 = 32%  H2

Now to calculate % mass 

we have to calculate the mass of the components of mixture: (mass= mol x Molecular weight)

Before reaction:
m(CO) = n(CO) × MW(CO) = 1 mol × 28 g mol-1 = 28 g CO
m(H2O) = 1.5 mol × 18 g mol-1 = 27 g  H2O

total mass =27 +28 = 55,0

After reaction:
m(CO) = 0,2 mol × 28 g mol-1 = 5.6 g  CO
m(H2O) = 0.7 mol × 18 g mol-1 = 12.6 g   H2O
m(CO2) = 0.8 mol × 44 g mol-1 = 35.2 g    CO2
m(H2) = 0.8 × 2 g mol-1 = 1.6 g  H2

then because   the total mass is 28 + 27 =55 g we can calculate % m :

55 :5,6 =100 : X      X= 10 2. mass % of CO

55:12,6 = 100 :X     X=22 9. mass % of H O

55 : 35,2 = 100 : X  X= 64 0. mass % of CO2

55 : 1,6 = 100 : X     X= 2.9 mass % of H2

PROBLEM 4
An alloy consists of rubidium and one of the other alkali metals. A sample of 4.6 g of the alloy when allowed to react with water, liberates 2.241 dm3 of hydrogen at STP.
Problems:
4.1 Which alkali metal is the component of the alloy?
4.2 What composition in % by mass has the alloy?

Relative atomic masses of alkali metals are :
Ar(Li) = 7; Ar(Na) = 23; Ar(K) = 39; Ar(Rb) = 85.5; Ar(Cs) = 133

solution

The reaction is :

2 M + 2 H2O → 2 MOH + H2

we know that at STP 1 mol have a volume of 22,414 L then  moles of H2 developed from reaction are 0,1 mol H2

if 0,1 mol are developed then 2 x 01 mol of Metal M goes to reaction

Because we have 4.6 g of the alloy   we can calculate the mean MW   (MW = g/mol )

4,6/0,2 = 23 g/mol

if mean is 23 is impossible the alkali metal is Na or K  because only Na MW is 23 and  K MW  is 39 but the alloy have to be  23+ metal .

From data we can be sure the alkali metal is Li and also if mean MW is 23  only Rb can be  the Metal not Cs .

We know that mol Rb + mol Li = 0,2 and g Rb + g Li = 4,6  g but g Rb = mol Rb x MW Rb      and g Li = mol Li x Mw Li

from this data we can write

n(Rb) + n(Li) = 0.2 mol
m(Rb) + m(Li) = 4.6 g
n(Rb) M(Rb) + n(Li) M(Li) = 4.6 g

but mol Li = 0,2 – molRb then
n(Rb) M(Rb) + (0.2 – n(Rb)) M(Li) = 4.6
n(Rb) . 85.5 + (0.2 – n(Rb)) × 7 = 4.6

mol Rb are :
n(Rb) = 0.0408 mol  then    n(Li) = 0.1592 mol       g Rb= 0,0408 x 85,5= 3,49    g Li = 0,1592  x 7  = 1,114

then if in 4,6 g there are 3,49 g then in 100 there are X

% Rb = 3,49 x 100/ 4,6 g  = 75,8

% Li = 1,114 /4.6 = 24,2

PROBLEM 5
An amount of 20 g of cooper (II) oxide was treated with a stoichiometric amount of a warm 20% sulphuric acid solution to produce a solution of copper (II) sulphate.
Problem:
How many grams of crystalline copper(II) sulphate (CuSO4 . 5 H2O) have crystallised when the solution is cooled to 20 °C?
Relative atomic masses: Ar(Cu) = 63.5; Ar(S) = 32; Ar(O) = 16; Ar(H) = 1
Solubility of CuSO4 at 20 °C: s = 20.9 g of CuSO4 in 100 g of H2O.

solution

the reaction is

CuO + H2SO4 → CuSO4 + H2O

mol CuO = 20/ 79,5 = 0,2516 g

mol H2SO4 = mol CuSO4 = mol CuO = 0,2516

the mass of CuSO4 solution  obtained is   mass Cuo + mass H2SO4 (20%)   but mass H2SO4 = 0,2516 x 98 =24,66 g  /0,2 = 123,3  then

m(solution CuSO4) = 143.28 g

Mass fraction of CuSO4:
a) in the solution obtained:

W(CuSO4) = m(CuSO4) /m(solution CuSO4)   = n  (CuSO4 )x m (CuSO ) /m(solution CuSO4) = 0,28 g

b) in saturated solution of CuSO4 at 20 °C:

W(CuSO4 ) =20 9. g/ 100+20 9. g = 0 ,173

c) in crystalline CuSO4 . 5 H2O:

W(CuSO4) = M(CuSO4) / M (CuSO4.5H2O)= 0,639

Mass balance equation for CuSO4:
0.28 m = 0.639 m1 + 0.173 m2
m – mass of the CuSO4 solution obtained by the reaction at a higher temperature.
m1 – mass of the crystalline CuSO4 . 5H2O.
m2 – mass of the saturated solution of CuSO4 at 20 oC.

0.28 × 143.28 = 0.639 m1 + 0.173 × (143.28 – m1)

0.28 × 143.28 = 0.639 m1 + 0.173 × (143.28 – m1)
m1 = 32.9 g
The yield of the crystallisation is 32.9 g of CuSO4 . 5H2O.

PROBLEM 6
Oxide of a certain metal contains 22.55 % of oxygen by mass. Another oxide of the same metal contains 50.48 mass % of oxygen.
Problem:
1. What is the relative atomic mass of the metal?
solution

If the first oxide for 100 parts we have 22,5 parts of O then parts are Me is 77,5 .The other oxide have 50,48 parts of O then Me is 49,52 parts.

the first oxide is    Me2Ox  then   2/x = 0,225/16 = 0,775/Me      2/X = 54,95 / Me 

the second oxide  Me2Oy then

2/y= 50,48 /16= 49,52 / Me       2/y = 15,695/ Me

then   y/x  = 54,95/ 15,695= 3,5   or y/x= 7/2

X=2 and      then 2/X= 54,95       2/2 = 54,95/Me      Me= 54,95 x 1 = 54,95 

the metal is  Mn (MW = 54,93) and the first oxide is MnO   and the second is   Mn2O7

THE FOURTH INTERNATIONAL CHEMISTRY OLIMPIAD  1972, MOSCOW

PROBLEM 1
A mixture of two solid elements with a mass of 1.52 g was treated with an excess of hydrochloric acid. A volume of 0.896 dm3 of a gas was liberated in this process and 0.56 g of a residue remained which was undissolved in the excess of the acid. In another experiment, 1.52 g of the same mixture were allowed to react with an excess of a 10 % sodium hydroxide solution. In this case 0.896 dm3 of a gas were also
evolved but 0.96 g of an undissolved residue remained. In the third experiment, 1.52 g of the initial mixture were heated to a high temperature without access of the air. In this way a compound was formed which was totally soluble in hydrochloric acid and 0.448 dm3 of an unknown gas were released. All the gas obtained was introduced into a one litre closed vessel filled with oxygen. After the reaction of the unknown gas with oxygen the pressure in the vessel decreased by approximately ten times (T = const).
Problem:
1.1 Write chemical equations for the above reactions and prove their correctness by calculations.
In solving the problem consider that the volumes of gases were measured at STP and round up the relative atomic masses to whole numbers

solution

1.1 a) Reaction with hydrochloric acid:
1.52 g – 0.56 g = 0.96 g of a metal reacted and 0.896 dm3 of hydrogen (0.04 mol) were formed.

Combining mass of the metal: 11.2 x 0,96 / 0,896=  12  g

This may be C or

2 x 12 =24 =MW Mg

or 3 x 12 =36  MW Cl

Possible solutions:

Relative  atomic mass Me                                Oxidation number                       Element          Satisfying?

12                                                                                          I                                   C                    No

24                                                                                         II                                  Mg                 Yes

36                                                                                         III                                 Cl                    No

Reaction: Mg + 2 HCl → MgCl2 + H2

infact   24 : 2,008 = 0,96 :X      X =0,08 H2    or 0,04 H

b) Reaction with sodium hydroxide:
1.52 g – 0.96 g = 0.56 g of an element reacted, 0.896 dm3 (0.04 mol) of hydrogen were formed.

0.56 combining mass of the metal: 11.2 x  0,56 / 0,896 = 7 g

Relative  atomic mass Me                                Oxidation number              Element        Satisfying?

7                                                                                         I                                      Li                   No
14                                                                                      II                                     N                    No
21                                                                                     III                                    Ne                  No

28                                                                                    IV                                      Si                   Yes

Reaction: Si + 2 NaOH + H2O → Na2SiO3 + 2 H2

c) Combining of both elements:
0.96 g Mg + 0.56 g Si = 1.52 g of silicide MgxSiy

0,96/24 =0,04       0,56 / 28 =0,02

dividing for 0,2  we have     x=2    y=1     then  Mg2 Si

silicide: Mg2Si

d) Reaction of the silicide with acid:
Mg2Si + 4 HCl → 2 MgCl2 + SiH4

n(Mg2Si) =1,52/76= 0,02 mol

n(SiH4) = 0,448/ 22,4 = 0,02 mol

e) Reaction of silane with oxygen:
SiH4 + 2 O2 → SiO2 + 2 H2O
V = 1 dm3
On the assumption that T = const:  p2=n2/n1P1

n1(O2) = 1/22,4 = 0,0446 mol

Consumption of oxygen in the reaction: n(O2) = 0.04 mol
The remainder of oxygen in the closed vessel:
n2(O2) = 0.0446 mol – 0.04 mol = 0.0046 mol

p = 0,0046/0,0446 p1= 0,1 P1

 

 

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