Mole Calculations, also commonly known as Mole Concepts & Chemical Calculations had been identified by students and educators alike, to be one #1 Killer Topic in GCE ‘O’ Levels Chemistry, IP Chemistry, IB Chemistry and IGCSE Chemistry.
Recently, we have seen more students asking us to discuss more in this chemistry blogsite. One of my chemistry student, who is committed to do well in the october’s GCE ‘O’ Levels Chemistry Examinations, had sent me a series of questions related to Mole Calculations.
They are questions that are not really difficult to comprehend and score, but you must have strong essential concepts & a good strategy to solve it. Let’s check out the questions:
Question 1:
A mixture of 8g of hydrogen gas and 8g of oxygen gas is burnt and the rxn is represented by the equation below. What is the mass of water formed?2H2(g) + O2 (g) –> 2 H2O (l)
A) 9g
B) 18g
C) 36g
D) 72g
&
Question 2:
An element of E forms a hydride with the formula EH3 which contains 90% of E by mass. What is the relative atomic mass of E?A) 27
B) 30
C) 87
D) 90
&
Question 3:
Which ion is present in highest concentration in a 2 mol/dm3 sodium sulfate solutionA) Na+
B) SO4 2-
C) H+
D) OH-
&
Question 4:
How many chloride ions are in contact with each sodium ion in a crystal lattice of sodium chloride?A 1
B 2
C 4
D 6
If you are a student that will be taking GCE ‘O’ Level Chemistry Exams or other related end-of-year Chemistry Examinations, the above sets of questions will be interesting for you to solve.
PS: Leave your suggested answers and comments in “Leave A Reply” below.
PPS: If you are not strong in Mole Calculations, and would like to have a quick-fix, you might want to consider the upcoming Mole Calculations Mastery Workshop.
Until the next post! See ya!
Related Articles:
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- O Level Chemistry: Mole Calculations – Concentrations of Solutions
- O Level Chemistry: Challenging Mole Calculation Questions
- O Level Chemistry: Challenging Mole Calculation / Electrolysis Questions
- O Level Chemistry: Challenging Mole Calculation / Electrolysis Questions
1) A. Must take the mole of limiting reagent, O2.
2) A.
3) D.
4) A.
Reply
1)D
2)A
3)B
4)A
Reply
mike Reply:
September 8th, 2009 at 8:38 pm
1)D
2)A
3)B
4)A
Reply
1)A
Remember, O2 is the limiting reactant. There is a common mistake by some students whereby they calculate the “No. of moles of H2″ and applied ratio directly, thinking that it is 4 moles.
Since O2 = LR,
Moles of O2 = 8/32 = 0.25
Moles of H2O reacted = 0.25 * 2 = 0.5
Mass of H20 = 0.5 * 18 = 9g
2)A
Relative Atomic Mass of H = 1
thus for H3 = 3 * 1 = 3 (which is 10%)
Relative Atomic Mass of E = 90% = 3 * 9 = 27
3)A
2 mol/dm3 = 2 moles of Na2SO4
1 Mole of Na2SO4 dissociates into 2 Na(+) and 1 SO4(2-) ions
this 2 moles of Na2SO4 = 4 moles of Na(+) and 2 moles of SO4(2-)
Thus, there is more Na+ ions
Common mistake here is people consider H+ or OH-.
Take note that in solvation, it is the WATER MOLECULES that surround the ions. (Not H+ or OH- ions). In other words, conc. of H+ and OH- should be 0.
Many students are confused due to the way they write the half equations of H+ and OH- ions in electrolysis for aqueous solutions. In reality, it is the H2O molecules that undergo electrolysis first, thus forming the ions. [Please ignore if its too confusing]
HINT: Anyway, even if u assume water = H+ and OH- instead of H2O(l),
you should atleast be able to tell that each mole of water = 1H+ and 1OH-.. thus the conc. if both ions would be the same… how can you have 2 correct answers in a MCQ question?
4)D
Go take a look at the crystal lattice diagram. Look at the Na+ or Cl- ions IN THE MIDDLE of the lattice. There should be 6 of the other ions next to it. 1 Top, 1 Bottom, 1 Left, 1 Right, 1 Front, 1 Back.
Common mistake is to take NaCl = Na+ and Cl- and conclude the answer to be 1.
Note: if each Na+ has 6 accompanying Cl-
and each Cl- has 6 accompanying Na+
the ratio is stilll 1:1
Reply
huaan Reply:
September 26th, 2009 at 2:00 am
u seem familiar. do i knw you?
Reply
JM Reply:
September 28th, 2009 at 7:35 pm
Nope i dont think so
Reply
waleed bin qasim Reply:
October 9th, 2009 at 11:03 pm
1.D
2.A
3.B
4.A
pliz grade sir
1) A
Firstly, calculate limiting reactant
Since we knw tat 2mole of H2 react with 1 mole of O2
4moles of H2, will react with 2 moles of O2.
Hence, We knw that oxygen is used up, since there are only 0.25 moles in the given reaction
Now we knw tat Oxygen is the limiting reactant, therefore, we can say that Moles of water is 0.5
Mass Of Water = 9g
2) A
10% -> hydride
3/(3+x) =1/10
x=27
3) C
Sodium Sulphate is a salt, formed by Sodium and Sulphuric Acid.
It is an acidic salt, as sodium sulphate is Na2SO4.
We all knw tat acidic salt dissociate in water to form H+ ions.
Hence, the ions with the highest concentrations should probably be H+ ions, and neither Na ions nor SO4 ions.
4) C
Indeed, In a Crystal lattice. the ratio of sodium ions is 1:1, and that is why the formula unit is NaCl.
However,the question now states that “chloride ions are in contact with each sodium ion in a crystal lattice ” now tat ionic compound forms a giant ionic structure,
1 Na+ions should have 4 other Cl- ions.( It shall the same for chloride as well)
Reply
JM Reply:
September 28th, 2009 at 7:42 pm
For 3),
- Sodium Sulphate is neutral not acidic. It can be formed from H2SO4, or an acidic salt, NaHSO4.
- A way to define/tell acidity is pH. But what is pH? It is a measure of the concentration of H+ ions. You cant dessociate H+ from Na2SO4.
For 4),
- Please look at it in a 3 dimension manner. Or simple flip open your textbook and look at the CENTER ions then start counting.
Reply
huaan Reply:
October 2nd, 2009 at 1:34 am
wow:D thanks JMC:D
Reply
Can you tell me where i can find your book in pakistan
Reply
1)A
2)A
3)B
4)A
Reply
1) A
2)A
3)B
4)A
Reply
seriously guys. the ANS is A A A D. omg u guys need to buck up ur chem!
Reply
huaan Reply:
October 2nd, 2009 at 1:36 am
we are all learning here. dun crap.
Reply
D.I think it should be C b/c sodium ion is surrounded by 6 Cl ions
Reply
oh sorry there is a mistake it should be D(6)
Reply
FOR 4
C(6)
Reply
2m————-4g
x—————8
H2 =4m
2m—————32g
x—————-8g
O2=0.5 m so it is limiting reactant
32—————36
8g————–x
x =9
so ans for 1 is A
2) x+3/100 = 3/10
x = 27 so ans = A
3) i did not uderstand
4) D
Reply
me too
Reply