As mentioned in earlier post on Atoms, Molecules & Stoichiometry – there aren’t many differences in the Mole Calculations questions between GCE A-Level (Advanced Level) Chemistry & GCE O-Level (Ordinary Level) Chemistry, in terms of formulae and approach.
However, in GCE A-Level Chemistry (H1/H2), the examiners expect you to have a good grasp of Basic Chemistry knowledge i.e. topics in GCE O-Level Chemistry should be at your finger tips.
Take the following question that i recently gave it to my A-Level JC1 H2 Chemistry Weekly Class.
Question:
What is the mass of Zn obtained when 50g of Zinc Oxide is reduced by 50g of charcoal?
In this Mole Calculations question, students are expected to be able to write the balanced chemical equation based on the information given. Half of my students can’t write the balanced chemical equations! Because they have forgotten about the topic of Metals & Reactivity Series in GCE O-Levels!
Warning: Do not throw away your GCE O-Level Chemistry notes just yet. You might need it at times for quick revision.
Let’s take a look at the approach to solve this question.
Suggested Answer:
Recalling what we have learned in GCE O-Level Chemistry, when ZnO is reduced by charcoal (which is essentially Carbon) – we will obtain the Zn metal and Carbon Monoxide as the products.
ZnO (s) + C (s) –> Zn (s) + CO (g)
50g 50g
No. of moles of ZnO = Mass of ZnO / Mr of ZnO = 50 / (65.4 + 16) = 0.614 mol
No. of moles of C = Mass of C / Ar of C = 50 / 12 = 4.17 mol
Therefore, ZnO is the Limiting Reagent & C is the Excess Reagent
No. of moles of Zn formed = No. of moles of ZnO = 0.614 mol
[Limiting Reagent determines the amount of product(s) formed]
Therefore, Mass of Zn formed = Mole of Zn X Ar of Zn = 0.614 X 65.4 = 40.2g (3 sig. fig)
Hope the above example gives you an idea on some of the expectations you need when making transition from GCE O-Level Chemistry to GCE A-Level Chemistry.
Let’s try out one question below.
Quick Check 1:
Calculate the volume of CO2 produced (at r.t.p) by decomposing 15g of calcium carbonate.
PS: Try out the question and leave your answers in the comment section below.
volume= 6.25*10^-3
3.6 dm-3 of carbon dioxide produced.
3.6dm
CaCO3 (s) -> CO2 (g) + CaO (s)
nCaCO3 = 15/(12+3(16)+40.1) = 0.14985 mol
Since mole ratio of CaCO3:CO2 is 1:1, nCO2 = 0.14985 mol
At r.t.p., Vol of CO2 produced = 0.14985 x 24 = 3.5964 dm³ = 3600 cm³ (3sf)
no.of moles of CaCO3=15/100
=0.15mol
accrdn to eqn>>> CaCO3–>CaO+CO2
THERFRE,no.of moles of CO2=0.15mol as well!!
hence vol of CO2=0.15X24
=3.6dm cube
Job well done everyone! Answer is indeed 3.6 dm3 or 3600cm3 for the volume of CO2 produced.
Impt is to write out the thermal decomposition equation of CaCO3 to determine the molar ratio:
CaCO3(s) -> CaO(s) + CO2(g)
1 mol of CaCO3 produced 1 mol of CO2.
Cheers!
Keep learning and sharing!
Sean Chua 🙂