INTERNATIONAL CHEMISTRY OLYMPIAD (Praga 1968)
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.
1.1 What is the composition of the initial gaseous mixture?
1.2 How chlorine, hydrogen, and hydrogen chloride are produced?
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 %
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
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
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.
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
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
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.
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
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 02 for 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 %
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.)
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 10–3mol
we know also that volume1 x M1 = volume2 x M2 then
moles of acid are m= 3.17 x 10–3mol
But grams /MW = moles then 0,19 / 3.17 x 10–3mol = 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
INTERNATIONAL CHEMISTRY OLYMPIAD ( Katowice 1969)
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.
What volumes of hydrogen and oxygen were obtained at a temperature of 20 °C and a pressure of 101 325 Pa?
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
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.
2.1 What are the stoichiometric formulas of the compounds?
2.2 Write the corresponding chemical equation
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 ↑
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
What is the molecular formula of the unknown hydrocarbon if volumes of gases were measured at standard temperature and pressure (STP) conditions?
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
Calcium carbide and water are the basic raw materials in the production of:
b) acetic acid
c) ethylene and polyethylene
d) vinyl chloride
Give basic chemical equations for each reaction by which the above mentioned compounds can be obtained.
Basic reaction: CaC2 + 2 H2O = Ca(OH)2 + C2H2
From acetylene can be obtained:
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->
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
3 CH≡ CH 400 -500 °C —-> C6H6 benzene
INTERNATIONAL CHEMISTRY OLYMPIAD ( Budapest, 1970)
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.
What is the structural formula of the gas (compound)?
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
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).
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.
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
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?
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)
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
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
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.
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
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
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.
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.
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.
Oxide of a certain metal contains 22.55 % of oxygen by mass. Another oxide of the same metal contains 50.48 mass % of oxygen.
1. What is the relative atomic mass of the metal?
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
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).
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
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
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
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