Caustic concentration evaporator problem

Question - 22 25 tons of caustic solution having 10 % concentration was fed to an evaporator . the concentrated liquor leaving the evaporator contains 50% NaOH , 2 % Nacl and 48 % H2O .

Calculate a - the pounds of water evaporated per hour

b- the pounds of salt precipitated per hour

c- pounds of concentrated liquor produced per hour
solution -

PART A- as the NaOH is the only constituent passing through the evaporator unchanged, it is convenient to base the concentrations of the H2O and Nacl on the NaOH.
INPUT OF NaOH = 25 *2000 * 10/100 = 5000 LB/HR
WATER REMOVED = 5000 *( 80/10 - 48/50) = 5000*( 8-0.96 ) = 35200 LB/HR

PART B-
AMOUNT OF SALT = 5000 (10/10- 2/50 ) = 5000 ( 1-0.098) = 4800 LB/HR
PART -C
THE RATIO OF CONCETRATED LIQOUR = 100/50
AMOUNT OF Na OH = 5000 *100/50 =10000 LB/HR
OR FEED BALANCE = 2000*25 = 50000 LB/HR
SO , CONCENTRATED LIQUO R = 50000 - (35200 +4800 ) = 10000 LB/HR