A mixture contains only NaCl and Al 2 (SO 4 ) 3 . A 3.47-g sample of the mixture

Question

A mixture contains only NaCl and Al2(SO4)3. A 3.47-g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of Al(OH)3. The precipitate is filtered,dried and weighed. The mass of the precipitate is 0.739 g. What is the mass percent of Al2(SO4)3 in the sample?

Also.....

Complete and balance each acid-base reaction with its specific states such as (aq) or (s) if needed:

(a) H2SO4(aq) + K OH(aq) --->

Contains two acidic hydrogens

(b) H2SO4(aq) + Fe(OH)3(s) --->

Contains two acidic hydrogens

(c) H2S(aq) + Sr(OH)2 --->

Contains two acidic hydrogens

(d) H4P2O7(aq) + Na OH(aq) --->

Contains four acidic hyderogens

Also......

The units of parts per million(ppm) and parts per billion(ppb) are commonly used by enviornmental chemists. In general, 1ppm means 1 part of solute for every 106 parts of solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions:

(a) 6.5 ppb Hg in H2O

(b) 2.7 ppb CH Cl3 in H2O

(c) 62.0 ppm As in H2O

(d) 0.51 ppm DDT (C14H9Cl5) in H2O

Explanation / Answer

Al2(SO4)3 + 6NaOH --> 2Al(OH)3 + 3Na2(SO4)

molecular weight of Al2(SO4)3 = 2 X 27 + 3 X 32 + 2X 16 =342

Molecular weight of Al(OH)3 =27+3*17= 78

As per the reaction

78 gms of Al(OH)3 gets precipitated from 342 gms of Al2(SO4)3

0.739gms of Al(OH)3 requires 0.739*342/78=3.24gms

Mass of sample =3.47 gms

Mass of Al2(SO4)3= 3.24 gms,Mass % of Al2(SO4)3 =100*(3.24/3.47)=93.37%

H2SO(aq)+ 2KOH (aq) -----> K2SO4(aq)+ 2H2O(l)

2 Fe(OH)3 (s) + 3 H2SO4(aq) Fe2(SO4)3(aq) + 6 H2O(l)

H2S(aq) + Sr(OH)2(aq)----. SrS(aq)+ 2H2O(l)

H4P2O7 (aq) + 2 NaOH (aq)= 3 H2O (aq) + 2 NaPO3(aq)

a) 6.5ppb means =6.5 parts in 1,000,000,000 = 6.5 g/ 1,000,000, 000 gm = 6.5*106mg/ 1,000,000,000 liter ( since density of such dilute solution =1g/cc)= 6.5 g/109cc=6.5g/106lter==6.5*10-6 gm/liter

atomic weight of Hg =201

Molarity =gmoles/liter of solution

Moles of Hg/liter of the solution =(6.5/201)*10-6 gmoles/liter=3.23*10-8M

b) 2.7ppb of CHCl3 in H2O similarly =7*10-6gm/liter Molecular weight of CHCl3= 12+1+3*35.5=119.5

Molarity of CHCl3 =(7/119.5)*10-6 gmoles/liter=0.058577*10-6 gmoles/liter

c) 62ppm As =(62 g/106 gms = 62gm/ 106cc = 62gms/103liter (1000cc =1liter)

atomic weight of As=75

Molarity = (62/75)*10-3 gmoles/liter= 8.3*10-3gmoles/liter

d) 0.51ppmDDT = 0.51*10-3gm/liter

Molecular weight of DDT (C14H19Cl5)= 12*14+9+5*31.5=354.5

Molarity= 0.51*10-3/354.5=1.44*10-6M

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