422
Reactions Prediction and Synthesis
SOLUTION:
The mixture is treated with just enough water to dissolve all of the more soluble
component (KNO 3 ) at 100 0°C Part of the NaCl also is dissolved, but a large
fraction of it remains as a residue of pure NaCl, which is separated by filtration at
100 0°C The filtrate is diluted with enough water to dissolve all of the remaining
NaCl at 0 0°C, and then it is cooled to 0 0°C where a precipitate of pure KNO 3 is
obtained and filtered off at 0 0°C
Step I Treat the mixture with enough water to dissolve 500 g of KNO 3 at
100 0°C
Vol of H 2 O required = (500 g KNO 3 ) ( ™ ^J^^} =
202 ml H zO
\ 247 g K.NO3/
/ 39 1 e NaCl\
Wt of NaCl dissolved in 202 ml = (202 ml H 2 O) I , ' , ' „ = 79 0 g NaCl
\ 100 ml H^O/
The undissolved NaCl (500 g - 79 0 g = 421 g pure NaCl) is filtered off
Step 2 Dilute the filtrate to the volume required to dissolve 79 0 g NaCl at
00°C
Vol of H 2 O required = (79 0 g NaCl) (
10
°
ml U ^\ = 221 ml H 2 O
\ 35 7 g NaCl/
221 ml - 202 ml = 19 0 ml water added to filtrate
Wt of KN0 3 dissolved in 221 ml = (221 ml H 2 O) (
1 3 J ,
3
= 29 4 g KN0 3
\ 100 ml H 2 O /
The undissolved KNO 3 (500 g - 29 4 g = 470 6 g pure KNO 3 ) is filtered off
PROBLEM:
What weights of pure salts can be recovered by one more cycle of fractional
crystallization
9
SOLUTION.
You have a solution that contains 29 4 g KNO 3 and 79 0 g NaCl The simplest thing
to do is evaporate off all the water, then repeat the cycle of the previous problem
Step 3 Treat the solid mixture with enough water to dissolve 29 4 g KN0 3 at
100 0°C
Vol of H 2 0 required = (29 4 g KNO 3 )
= ] ! 9 ml H >>
O
/ 39 1 e NaCl\
Wt of NaCl dissolved in 11 9ml = (11 9 ml H 2 O) nn
e
, T ^ I = 4 7 g NaCl
\ 100 ml H 2 O/
The undissolved NaCl (79 0 g - 4 7 g = 74 3 g pure NaCl) is filtered off
Step 4 Dilute the filtrate to the volume required to dissolve 4 7 g NaCl at 0 0°C
Reactions Prediction and Synthesis
SOLUTION:
The mixture is treated with just enough water to dissolve all of the more soluble
component (KNO 3 ) at 100 0°C Part of the NaCl also is dissolved, but a large
fraction of it remains as a residue of pure NaCl, which is separated by filtration at
100 0°C The filtrate is diluted with enough water to dissolve all of the remaining
NaCl at 0 0°C, and then it is cooled to 0 0°C where a precipitate of pure KNO 3 is
obtained and filtered off at 0 0°C
Step I Treat the mixture with enough water to dissolve 500 g of KNO 3 at
100 0°C
Vol of H 2 O required = (500 g KNO 3 ) ( ™ ^J^^} =
202 ml H zO
\ 247 g K.NO3/
/ 39 1 e NaCl\
Wt of NaCl dissolved in 202 ml = (202 ml H 2 O) I , ' , ' „ = 79 0 g NaCl
\ 100 ml H^O/
The undissolved NaCl (500 g - 79 0 g = 421 g pure NaCl) is filtered off
Step 2 Dilute the filtrate to the volume required to dissolve 79 0 g NaCl at
00°C
Vol of H 2 O required = (79 0 g NaCl) (
10
°
ml U ^\ = 221 ml H 2 O
\ 35 7 g NaCl/
221 ml - 202 ml = 19 0 ml water added to filtrate
Wt of KN0 3 dissolved in 221 ml = (221 ml H 2 O) (
1 3 J ,
3
= 29 4 g KN0 3
\ 100 ml H 2 O /
The undissolved KNO 3 (500 g - 29 4 g = 470 6 g pure KNO 3 ) is filtered off
PROBLEM:
What weights of pure salts can be recovered by one more cycle of fractional
crystallization
9
SOLUTION.
You have a solution that contains 29 4 g KNO 3 and 79 0 g NaCl The simplest thing
to do is evaporate off all the water, then repeat the cycle of the previous problem
Step 3 Treat the solid mixture with enough water to dissolve 29 4 g KN0 3 at
100 0°C
Vol of H 2 0 required = (29 4 g KNO 3 )
= ] ! 9 ml H >>
O
/ 39 1 e NaCl\
Wt of NaCl dissolved in 11 9ml = (11 9 ml H 2 O) nn
e
, T ^ I = 4 7 g NaCl
\ 100 ml H 2 O/
The undissolved NaCl (79 0 g - 4 7 g = 74 3 g pure NaCl) is filtered off
Step 4 Dilute the filtrate to the volume required to dissolve 4 7 g NaCl at 0 0°C