Recently, i received several emails and comments from readers (students) to share more challenging Mole Calculations (or Mole Concepts) questions in this chemistry blog.? So here we go………..
Question:
What would be the concentration of the hydrochloric acid produced if all the hydrogen chloride gas from the reaction between 50 g of pure sulfuric acid and an excess of sodium chloride was collected in water, and the solution made up to a volume of 400 cm3 of water?
Based on equation: NaCl(s)?? +?? H2SO4(aq)??? –> ? NaHSO4(s)?? +?? HCl(g)
Now, try it out and leave your suggested answer (and possibly your solution/steps) in the comment section below.
The best way to learn is NOT by reading or looking, instead it is by Taking Actions.
I look forward to seeing you answers. Cheers! Enjoy! =)
NaCl + H2SO4 -> NaHSO4 + HCL
n•(H2SO4) =?
m(H2SO4)= 50g
Mr(H2SO4)= 98g/mol
therefore n= m/Mr
n= 50g/98g/mol = 0.51mol.
according to the mole ratio we have 1mole of each compound.
therefore n•(H2SO4) : n•(HCL)
1 : 1
0.51 : x
therefore n(HCl) = 0.51mol
conc=n• / Volume
conc= 0.51/ (400cm3/1000)
conc=0.51mol/0.4dm3
conc= 1.28mol/dm3
I didn’t understand the last two steps :O
I feel it is really fun for you guys to early tackle questions,as for me I think I’m still far away,even if I have much passion.
Hi Abel,
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can you please be my personal tutor
Mr or H2SO4 = 98
98g H2SO4 = 1 MOL
50g H2SO4 = 0.51 MOL
1 MOL H2SO4 = 1 MOL HCL
0.51 MOL H2SO4 = 0.51 MOL HCL
NO OF MOL HCL IN 40OCM^3 WATER = 0.51 MOL
NO OF MOL HCL IN 100CM63 WATER = 1.275 MOL
THEREFORE CONCENTRATION = 1.275mol/dm^3
my method is rather kind of long but you can understand well right?
1.28 mol/dm3 (to 3 sf)
@Saud – you are correct, don’t worry.
its because the concentration is in mol./dm3, but the volume of water given is in cm3 so we divide it by 1000 to convert it into dm3. =D
1.28 mol/dm3
i don’t get that why others have divided 0.51 with 400/1000 …… ??
guide me guys…………..
yeah i agree with the majority….
its 1.28mol/dm3
mass of sulfuric acid = 50g
relative formula mass of sulfuric acid = 98
so,
moles of sulfuric acid = mass/molar mass
= 50/98
= 0.51 moles
keeping in view the stoichometric ratios of the reactants and the products we get that one mole of sulfuric acid results one mole of HCl…
given also that the solution was made upto 400 cm3
that means that the moles of HCl in 400cm3 are 0.51
how much will these be in 1000cm3 i-e 1dm3
that is
=1000/400 x0.51
=1.275 moles in 1dm3
approximating it to 2 decimal places
concentration of HCl = 1.28 mol/dm3
🙂
What would be the concentration of the hydrochloric acid produced if all the hydrogen chloride gas from the reaction between 50g of pure sulfuric acid and an excess of sodium chloride was collected in water, and the solution made up to a volume of 400 cm3 of water?
Based on equation: NaCl(s) + H2SO4(aq) ?> NaHSO4(s) + HCl(g)
mole=50/(2+32+(16×4))=0.5102
mol/dm3=0.5102/(400/1000)=1.28mol/dm3
It’s 1.28 mol/dm3
27.40 mol/dm3
50/98=0.5102
0.5102/(400/1000)= 1.28mol/dm3