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?
Answers
Answered by
3
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
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
Similar questions