Finding volume. (Mole concept- Chemistry)
-
Question:
The equation for the thermal decomposition of sodium hydrogen carbonate is:
2NaHCO3 (s) = Na2CO3 (s) + H2O (g) +CO2 (g).
Find the volume of carbon dioxide, evolved measured at s.t.p.
________________________________________________________
No. of moles of NaHCO3 = 21/(23+1+12+48) = 0.25mol.
No. of moles of CO2 = 0.25/2 = 0.125mol.
How to find volume?
I only know of Mol divided by Mol/dm^3, which seems to not fit this question's demands.
Would appreciate it very much if someone could share with me their answers and method for this particular question.
-
At standard temperature (zero deg C) and pressure (1 atm), the molar volume (ie. volume of one mole) of any gas (assuming ideal gas behaviour) is approximately 22.414 (or 22.4 to 3 sig fig) dm3.
Hence sample volume = no. of moles x molar volume = 0.125 x 22.414 = 2.80 dm3.
-
I dont understand this... This topic... I need to study. Quite complicated. Any advice? Will I be able to figure it out from textbook?
-
Originally posted by xXBlack_RebelXx:
I dont understand this... This topic... I need to study. Quite complicated. Any advice? Will I be able to figure it out from textbook?
u sec 3 or what? if sec 3.. textbook isnt much use. ive tried and i dun really understand much from the tb. even after i understood, its easily forgotten luhh. -
if u cant understand the one mole of ideal gas=22.4dm3(approx)
try looking at it this way
for an ideal gas
Pv=nRT
since its s.t.p, for all cases of ideal gas in s.t.p
V=n[(RT)/P]
since R,T and P are constants let [(RT)/P]=k where k is a constant
hence v=kn
if u calculate k=22.4 approx
thus we come to the conclusion that for ideal gas at s.t.p 1mol has a vol of 22.414 dm3
hope i am of help, doing things the long method helps you understand better rather than just memorising things like 1mol= 22.4dm3
-
I dont understand a thing... is this Pure Chem?
-
yes this is sec3 chem on mole concept..
so basically 1dm3 = 1000cm3 of either gas or liquid?
and what i've learnt is only r.t.p...so we use 24dm3
thanks for the prompt replies
-
Originally posted by xXBlack_RebelXx:
I dont understand a thing... is this Pure Chem?
Sounds like Black_Rebel could certainly benefit from some good tuition. In the meanwhile, feel free to keep asking questions on anything you don't understand, here.
And donkhead333, 1 litre = 1dm3 = 1000cm3 = 1000ml, for both gases and for solutions.
Also, 1m3 = 1000dm3 = 1,000,000cm3. Yes, for both gases and for solutions.
-
a second question (why do i keep receiving questions that i've never seen before)
question:
deuterium is the name given to the isotope of hydrogen that contains one proton and one neutron in its nucleus. the compound D2O (deuterium) is called "heavy water".
a) calculate the Mr of deuterium. why is the name "heavy water" appropriate for this molecule?
b) heavy water has the same reactions as normal water, but its reactions are usually slower. explain this difference in reaction rate.
thanks again lol
-
Originally posted by donkhead333:
a second question (why do i keep receiving questions that i've never seen before)
question:
deuterium is the name given to the isotope of hydrogen that contains one proton and one neutron in its nucleus. the compound D2O (deuterium) is called "heavy water".
a) calculate the Mr of deuterium. why is the name "heavy water" appropriate for this molecule?
b) heavy water has the same reactions as normal water, but its reactions are usually slower. explain this difference in reaction rate.
thanks again lol
a) calculate the Mr of deuterium. why is the name "heavy water" appropriate for this molecule?
The relative atomic mass or molar mass of a deuterium atom (D) is 2 or 2g.
The relative molecular mass or molar mass of a deuterium molecule (D2) is 4 or 4g.
The relative molecular/formula mass or molar mass of heavy water (D2O) is 20 or 20g.
Heavy water, D2O has a molar mass of 20g, which is heavier than regular water, H2O, which has a molar mass of 18g.
b) heavy water has the same reactions as normal water, but its reactions are usually slower. explain this difference in reaction rate.
The rate (determining step) of chemical reactions involving heavy water, will be slightly lower or slower, than the rate (determining step) of chemical reactions involving regular (lighter) water; since the heavier the mass (and hence greater the density) of reactant particles (atoms or molecules or ions), the slower the movement of the reactant particles; hence at the same temperature, the reaction pathway involving the heavier particles will have a lower percentage of particles with energy equal or exceeding activation energy required for the reaction, as compared with the reaction pathway involving the lighter particles, and hence a differential rate of effective collisions of reactant particles per unit time, results; and therefore a difference in the rate of reaction will be observed (between the reaction pathways involving heavier particles versus lighter particles).