0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas burns completely in O2 to form H2O and 72 cm3 of CO2 only. The volume of CO2 was measured at room temperature and pressure.
What could be the formula of X?
A Al2C3 B Al3C4 C Al4C3 D Al5C3
I have come up with the following solution:
X + H2O gives Al2O3 + CxHy. CxHy + O2 gives CO2 and H2O
72cm3
Moles of CO2 = 72/24000 = 0.003
If we get the number of moles of X then by using the given mass of X (0.144g) we can calculate the Mr of X and the formula can be made. For this I must know what is CxHy.
I have calculated the moles of CO2 and then i am stuck...........what to do next?
Originally posted by hoay:0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas burns completely in O2 to form H2O and 72 cm3 of CO2 only. The volume of CO2 was measured at room temperature and pressure.
What could be the formula of X?
A Al2C3 B Al3C4 C Al4C3 D Al5C3
Experimentally, moles of C present = 3x10^-3
Molar mass of AlxCy = 27x + 12y
Moles of AlxCy present = sample mass / molar mass = 0.144 / (27x + 12y)
By stoichiometry, moles of C present = moles of (AlxCy) x (y) = y (0.144 / (27x + 12y))
Which can be equated to 3x10^-3 found earlier.
Simplifying the mathematical expression, we obtain x/y = 4/3.
Hence the answer is C, aluminium carbide.
How can we deduce the moles of AlxCy as 0.003 if we don't know the stoichiometric ratios of AlxCy and CxHy!! Secondly we don't know what is CxHy! I did'nt get it...............Please explain.
Thanks
I've already given the full worked solution (except for the mathematical simplification process, which is more mathematics than chemistry).
My solution is already step-by-step and totally comprehensive. I suggest you print it out, and discuss the question and solution with your classmates. I'm sure you and your classmates can make full sense of it after a brief discussion.
Molar Volume = 24 dm^3 or 24 000 cm^3, It is the volume of gas 1 mole of any chemical occupy.
Question stated 72 cm^3 of CO2 produced from the previous reaction.
Therefore the number of moles of C = 72 cm^3 divided by 24, 000 cm^3 = 0.003 mol
When facing this type of question, it is not necesary to write up the chemical equation. This question tests you on your ability to deduce molecular formula through algebra and understanding on mole concept.
Q. When a mineral was heated in a Bunsen flame to constant mass, a colourless gas that turned lime water milky was evolved. The remaining solid was cooled and then added to aqueous hydrochloric acid. Vigorous effervescence was seen.
What was the mineral?
A aragonite, CaCO3
B artinite, MgCO3.Mg(OH)2.3H2O
C barytocalcite, BaCO3.CaCO3
D dolomite, CaCO3.MgCO3
The answer is C
In bunsen flame CaCO3 on decomposition gives CaO and CO2, while BaCO3 remains due to its high thernal stability. Then on reaction with HCl BaCO3 reatec to give CO2.
The same should have happened in case of D. Please explain.
Originally posted by hoay:Q. When a mineral was heated in a Bunsen flame to constant mass, a colourless gas that turned lime water milky was evolved. The remaining solid was cooled and then added to aqueous hydrochloric acid. Vigorous effervescence was seen.
What was the mineral?
A aragonite, CaCO3
B artinite, MgCO3.Mg(OH)2.3H2O
C barytocalcite, BaCO3.CaCO3
D dolomite, CaCO3.MgCO3
The answer is C
In bunsen flame CaCO3 on decomposition gives CaO and CO2, while BaCO3 remains due to its high thernal stability. Then on reaction with HCl BaCO3 reatec to give CO2.
The same should have happened in case of D. Please explain.
Amongst Ba2+, Ca2+ and Mg2+, the charge density of Ba2+ is the lowest, hence BaCO3 is most thermally stable.
Copper conatins Cu2+ as the basic particle but the formula of copper is written as Cu and not Cu2+?
Is it to distinguish it from the other compounds especially ionic such as CuCl2 in which there are no Cu atoms but Cu2+ ions are present.
Originally posted by hoay:Copper conatins Cu2+ as the basic particle but the formula of copper is written as Cu and not Cu2+?
Is it to distinguish it from the other compounds especially ionic such as CuCl2 in which there are no Cu atoms but Cu2+ ions are present.
Yes. Cu(s) implies Cu2+ ions together with their sea of delocalized valence electrons, ie. metallic copper.
The shape of BF3 is planar, where as CH4 is tetrahedral and not planar. What does the word planar suggests in this context? Does it means that all the atoms in BF3 including B lie at the same plane and in CH4 all the atoms do not lie in the same plane as we know its wedge nad dot form?
Please explain.
Originally posted by hoay:The shape of BF3 is planar, where as CH4 is tetrahedral and not planar. What does the word planar suggests in this context? Does it means that all the atoms in BF3 including B lie at the same plane and in CH4 all the atoms do not lie in the same plane as we know its wedge nad dot form?
Please explain.
Yes, you got the idea.
In BF3, there is a single plane (eg. a flat piece of glass) that all 4 atoms will align perfectly with. In CH4, no single plane (eg. a flat piece of glass) will be able to touch all 5 atoms.
So we can say that all the atoms in BF3 are co-planar while this is not the case in CH4?
Originally posted by hoay:So we can say that all the atoms in BF3 are co-planar while this is not the case in CH4?
Correct.
Co-planar implies exactly planar, while peri-planar implies near or almost planar. Both terms may be used interchangeably for 'A' level purposes.
The following two displacement reactions were carried out in a calorimeter with a heat capacity of 2000 J K-1.
Calculate the enthalpy change of reaction in both experiments.
The specific heat capacity of the copper(II) sulfate and silver nitrate solutions should be taken as 4.2 J K-1cm-3. You may assume that the mass of excess metal in both experiments is negligible.
In the first experiment, excess magnesium powder was added to 100 cm3 of a 1.00 mol dm-3 solution of copper (II) sulfate. The temperature rose from 19.5 °C to 41.2 °C.
Mg + CuSO4 MgSO4 + Cu
n = 0.1 mol
q = 100 x 4.2 x (41.2-19.5)
q = 9.11 KJ
enthalpy change = 9.11 / 0.1 = - 91.14 Kj/mol [ actual asnwer is -525 kj/mol]
In the second experiment, excess copper powder was added to 100 cm3 of a 0.500 mol dm-3 solution of silver nitrate. The temperature rose from 19.5 °C to 20.9 °C.
Cu + 2AgNO3 2Ag + Cu(NO3)2
n = 0.05
q = 100 x 4.2 x 1.4
q = 0.588 Kj/ mol
enthalpy change = 0.588 / 0.05 = -11.76 Kj/mol
Both are wrong. Where am i wrong.
(post deleted)
The following reaction was carried out in a calorimeter with a heat capacity of 2000 J K-1. The specific heat capacity of the solutions should be taken as 4.2 J K-1 cm-3. You may assume that the mass of excess metal in the experiments are negligible. In the first experiment, excess magnesium powder was added to 100 cm3 of a 1.00 mol dm-3 solution of copper (II) sulfate. The temperature rose from 19.5 °C to 41.2 °C. In the second experiment, excess copper powder was added to 100cm3 of a 0.50 mol dm-3 solution of silver nitrate. The temperature rose from 19.5 °C to 20.9 °C.
Calculate the enthalpy change of reactions in both experiments, as well as the expected enthalpy change, per mole of magnesium, for the reaction (not carried out experimentally) between magnesium and silver nitrate.
Solution :
The heat capacity of the entire system is the sum of the heat capacities for the individual components, ie. the calorimeter and the aqueous solution.
Total Heat Capacity = Heat Capacity of Water + Heat Capacity of Calorimeter
= (100cm3 x 4.2 J K-1 cm3) + (2000 J K-1)
= 2420 J K-1
Heat transferred in Experiment 1 = Total Heat Capacity x Change in Temperature
Q = (2420 J K-1) x (41.2 - 19.5)
Q = 5.2514 x 10^4 J
Heat transferred in Experiment 2 = Total Heat Capacity x Change in Temperature
Q = (2420 J K-1) x (20.9 - 19.5)
Q = 3.3880 x 10^3 J
Note that because the temperature rose, the reactions are exothermic, and therefore you must write the negative sign for the enthalpy change value.
Mg(s) + Cu2+(aq) ---> Mg2+(aq) + Cu(s)
Enthalpy change of 1st experiment in regard to either Mg or Cu2+
= Heat transferred / (moles of reactant of either Mg or Cu2+)
= - (5.2514 x 10^4 J) / (100/1000 x 1.0)
= - 5.2514 x 10^2 kJ/mol of Mg or Cu2+
Cu(s) + 2Ag+(aq) ---> Cu2+(aq) + 2Ag(s)
Enthalpy change of 2nd experiment in regard to Ag+
= Heat transferred / (moles of reactant of Ag+)
= - (3.3880 x 10^3 J) / (100/1000 x 0.5)
= - 6.7760 x 10^1 kJ/mol of Ag+
Enthalpy change of 2nd experiment in regard to Cu
= Heat transferred / (moles of reactant of Cu)
= - (3.3880 x 10^3 J) / (100/1000 x 0.5) / 2
= - 1.3552 x 10^2 kJ/mol of Cu
By Hess Law, enthalpy change of reaction between magnesium and silver nitrate
= (enthalpy change of reaction between magnesium and copper)
+ (enthalpy change of reaction between copper and silver nitrate)
= (-5.2514 x 10^2)kJ + (-1.3552 x 10^2)kJ
= -6.6066 x 10^2 kJ/mol of Mg