amount of substance multiple choice Flashcards
What is the minimum volume of 0.0500 mol dm–3 aqueous bromine needed to
react completely with 0.0200 g of buta-1,3-diene?
(Mr of buta-1,3-diene = 54.0)
A 7.40 cm3
B 14.8 cm3
C 29.6 cm3
D 67.5 cm3
Okay so your ratio will be 2Br : C4H6 (2:1) in the forwards direction.
You’re correct in your workings for the moles of buta-1,3-diene (approx 3.7x10^-4 moles). You then have to times this by 2 because you’re working backwards, giving you approx 7.4x10^-4 moles of Bromine.
The minimum volume is worked out using volume= moles/concentration i.e 7.4x10^-4/0.05. This gives you an answer of 0.0148. THIS IS STILL IN DM^3 so you have to times by 1000 to make it into cm^3.
When heated, a sample of potassium chlorate(V) (KClO3) produced 67.2 cm3 of
oxygen, measured at 298 K and 110 kPa
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
What is the amount, in moles, of potassium chlorate(V) that has decomposed?
The gas constant, R = 8.31 J K–1 mo
Use pV=nRT to find moles. pressure is in kpa, Volume is in dm3, R is 8.31, T is 298k (kelvin)
pv/RT = n
so (0.0672 x 110)/ 8.31 x 298 = moles
then divide it by 3 and multiply by 2 to get correct molar ratio (2:3) which gives you 1.99 x10-3 to 3sf
Q11.
A student rinsed the apparatus before starting an acid-base titration.
The results of the titration showed that the volume of acid added from the burette
was larger than expected.
Which is a possible reason for this?
A The conical flask was rinsed with water before
the titration.
B The walls of the conical flask were rinsed with
water during the titration.
C The pipette was rinsed only with water.
D The burette was rinsed only with water.
d
