An aqueous solution of sodium hydroxide contains 15% naoh by mass. it is desired to produce an 5% naoh solution by diluting a stream of the 15% solution with a stream of pure water. calculate the following: a) the ratios (g h, o/g feed solution) and (g product / g feed solution). b) determine the feed rates of 15% solution and diluting water needed to produce 2500 ib./min of the 5% solution.

Explanation:

We will assume that the flow rate of feed (F) = 100 g/min

(a)  The overall mass balance will be as follows.

           Total mass IN = Total mass OUT

as,          Feed + Pure water = Product

                100 + W = P

or,                P - W = 100 .......... (1)

Whereas NaOH mass balance will be as follows.

            Total NaOH mass IN = Total NaOH OUT

         NaOH in feed + NaOH in water = NaOH in product

                     P =

                         = 250 g/min

Now, we will put the value of P in equation (1) as follows.

                           P - W = 100

                          250 - W = 100

                          W = 250 - 100

                               = 150 g/min

Therefore,   =

                            =

                            = 1.5

      =

                            =

                            = 2.5

(b)   As it is given that 8% solution (P) = 2400 lb/min.

As it is calculated that the ratio of P and F () = 2.5

So,          F =

                 =

                 = 960 lb/min

Also,       = 1.5

                      W =

                          =

                          = 1440 lb/min

Therefore, we can conclude that the feed rate of 20% solution = F = 960 lb/min.

When diluted with water then W = 1440 lb/min.

since electrons are negative the answer is neagtaive