1. In which reaction does the smallest percentage change in volume occur?
A C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (l)
B 4NH3 (g) + 3O2 (g) -> 2N2(g) + 6H2O (l)
C CH4(g) + 2O2 (g) - > CO2 (g) + 2H2O (l)
D 2H2S (g) + SO2 (g) -> 3S (s) + 2H2O (l)
For me, I calculated the mole ratio in the equation. Eg. In reaction 1, we have 6 mols of gas reactants. Product is 3 mols of gas substance. So the change is 3 mols.
So do I use 3/6 X 100% = 50%, making this option having the smallest percentage change?
But for option D, this method is weird. With no gaseous products formed, do I find percentage change using 3/3 X 100 = 100%
2 8g of X2O3, an oxide of element X, contains 5.6g of X. How many moles of X does 5.6g of the element contain?
2.4/16 X (2/3)
2.4/16 X (1.5)
8/16 X (2/3)
8/16 X 1.5
I tried this for a long time. None of my answer fits the options. Not really sure of the concept. It'd be good if you could show me the step-by-step approach.
Thanks!
Talking about 2008 O's chem question, I have one up in my mind too.
There's this question on how you can show that sea water contains iodide ions besides using the displacement reaction.
The first thought that came up my mind was testing using acidified K2Cr2O7 because it can oxidise iodide ions to iodine. But the answer scheme gives me something like using the starch indicator and you observe drak blue-black stains. However, they are still iodide ions and not iodine so how can you turn the starch indicator blue black? Unless there's something in the indicator that can oxidise iodide ions (not sure about this, can someone confirm this for me?)
For question 1, your answer is correct. Your method should be valid. I'm also unclear about question 2.
Big thanks.
My answer for qn 2 would be the 1st 1.
2.4/16 gives u moles of oxygen and since the ratio is 2:3, 2/3 will give you the moles of X
Originally posted by dkcx:My answer for qn 2 would be the 1st 1.
2.4/16 gives u moles of oxygen and since the ratio is 2:3, 2/3 will give you the moles of X
Aaaa, how do you get 4/16?
I mean the mass of 4g for O2...
Originally posted by anpanman:
Aaaa, how do you get 4/16?I mean the mass of 4g for O2...
Its 2.4 and not 4 which can be obtained from the question since 5.6g is X so 2.4g must be O
For me, I calculated the mole ratio in the equation. Eg. In reaction 1, we have 6 mols of gas reactants. Product is 3 mols of gas substance. So the change is 3 mols.
So do I use 3/6 X 100% = 50%, making this option having the smallest percentage change?
But for option D, this method is weird. With no gaseous products formed, do I find percentage change using 3/3 X 100 = 100%
Correct, and yes.
Q2: No. of moles of O = [(8-5.6) / 16 ] = 0.15 mol
So... since 2X + 3O --> X2O3 (this equation is a simplified form of the idea I am trying to portray, no worries this is in A-Levels)
No. of moles of X = 0.10 mol
Ar of X = Mass/Moles = 56
So,
No. of moles of X in 5.6g of X = Mass/Molar Mass = 5.6/56 = 0.1 mol
So find the choice that gives 0.1, and that is the answer.
EDIT: dkcx method is a very efficient method, as the mass of X stated in both scenarios is the same. Take note that if both masses aren't the same, you'll have to go the long way to avoid confusion. If you can visualise fast enough though, then please, go ahead and do the shortcut to save time.
There's this question on how you can show that sea water contains iodide ions besides using the displacement reaction.
The first thought that came up my mind was testing using acidified K2Cr2O7 because it can oxidise iodide ions to iodine. But the answer scheme gives me something like using the starch indicator and you observe drak blue-black stains. However, they are still iodide ions and not iodine so how can you turn the starch indicator blue black? Unless there's something in the indicator that can oxidise iodide ions (not sure about this, can someone confirm this for me?)
Potassium dichromate (VI), K2Cr2O7, can oxidise a wide array of substances, so it might not be suitable to identify the presence of iodine alone. A colour change in K2Cr2O7 might not prove that iodide ions are oxidised and present, but it may be something else, like any common reducing agent perhaps?
Iodine and iodide ions are technically the same thing, just that iodine refers to the solid at room temperature, while iodide ions exist in compounds such as potassium iodide. The chemical compound you use to test for starch in the biology lab, do not be mistaken, that is not iodine alone. If I recall correctly, since they call in Aqueous Iodine, is Iodine + Potassium Iodide (to increase solubility of iodine, since iodine is sparingly soluble in water) + Water dilution
Originally posted by Audi:Talking about 2008 O's chem question, I have one up in my mind too.
There's this question on how you can show that sea water contains iodide ions besides using the displacement reaction.
The first thought that came up my mind was testing using acidified K2Cr2O7 because it can oxidise iodide ions to iodine. But the answer scheme gives me something like using the starch indicator and you observe drak blue-black stains. However, they are still iodide ions and not iodine so how can you turn the starch indicator blue black? Unless there's something in the indicator that can oxidise iodide ions (not sure about this, can someone confirm this for me?)
For question 1, your answer is correct. Your method should be valid. I'm also unclear about question 2.
Big thanks.
use easy method
Add dilute HNO3, aq AgNO3, if pale yellow ppt seen w/o efferversence of colourless gas, that is iodide
Originally posted by SBS n SMRT:use easy method
Add dilute HNO3, aq AgNO3, if pale yellow ppt seen w/o efferversence of colourless gas, that is iodide
The question asked for besides a displacement reaction. The one you have suggested produces silver iodide from an iodide compound and silver nitrate, which in this case, is a DOUBLE DISPLACEMENT REACTION.
Originally posted by Garrick_3658:The question asked for besides a displacement reaction. The one you have suggested produces silver iodide from an iodide compound and silver nitrate, which in this case, is a DOUBLE DISPLACEMENT REACTION.
which is not a displacement reaction... it is merely a double displacement reaction.. hence it is not a displacement reaction ? -.-
anyway. im sure that sbs ans is correct cause O lv qn nvr learn things like what is double reaction or what. we only learn the silver iodide and displacement of halogens only.. x.x
Originally posted by yiha093:which is not a displacement reaction... it is merely a double displacement reaction.. hence it is not a displacement reaction ? -.-
anyway. im sure that sbs ans is correct cause O lv qn nvr learn things like what is double reaction or what. we only learn the silver iodide and displacement of halogens only.. x.x
The question states besides using a displacement reaction. Adding HNO3 and AnNO3 is used to displaced the I- from the solution so it cannot be used to solve this question though its a possible method to test for I-.
The way i quoted is the best method to test for iodide ions, believe it or not, as textbook says it, plus, it is double displacement and not displacement, many candidates dun understand what is double displacement and what is displacement
Displacement is
Cl2(aq) + KBr (aq) --> KCl (aq) + Br2(aq)
Ionic eqn: Cl2 (aq) + 2Br-(aq) --> 2Cl-(aq) + Br2 (aq)
Double displacement
suppose iodide salt is KI
Ag(NO3)2 (aq) + KI (aq) --> KNO3(aq) + AgI (s) .........
ionic equation is Ag+ (aq) + I- (aq) --> AgI(s)
DOUBLE DISPLACEMENT =/= DISPLACEMENT
i hope this clarify