Draw the process and show the solution 1. A pure Na2B4O7 is to be recovered from

Question

Draw the process and show the solution

1. A pure Na2B4O7 is to be recovered from a solution containing 25 % Na2B4O7. The solution is heated to 100 oc where 75 % of the water is removed and the resulting solution is cooled gradually to 55 oc where Na2B4O7.10H2O crystals are produced. The crystals are separated by filtration and part of the liquor adheres to the filter cake. 5 g of the liquor adheres per 100 g crystals. The collected filter cake is sent to a dryer where all of the water was removed (including water of crystallization). Calculate the weight of the dry Na2B4O7 recovered. Solubility of Na2B4O7 at 55 oc is 14 g Na3B4O7 per 100 g water.

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Explanation / Answer

Ans: Molar mass of Na2B4O7= 201.22 & Molar mass of Na2B4O710H2O = 381.38

The mass of the solution: 100 g

The mass of dissolved Na2B4O7: m = 100 g · 25/100 = 25 g

The mass of crystals Na2B4O710H2O = x

x g Na2B4O710H2O contains: x 201.22/381.38 g Na2B4O7

The mass of Na2B4O7 remained in solution: 25- (x.201.22/381.38)

The mass of the solution: (100 - x) g

Since the concentration of the cooled solution is 14 %, the following relation exists:

(100 - x).14 = 25- (x.201.22/381.38)

x= 28.37 So, 28.37g of crystals Na2B4O710H2O obtained.

Now 100 g crystals = 5 g of the liquor adheres

So, 28.37 g crystals = (28.37 x 5)/100=1.4185 g of liquor adheres .

Total g of Na2B4O710H2O= 28.37+1.4185=29.7885 g

Now, According to molar mass 381.38 g of Na2B4O710H2O = 201.22 g Na2B4O7

                                           so, 29.7885 g of Na2B4O710H2O = (29.7885 x 201.22/381.38) g

                                                                                                 = 15.72 g of dry Na2B4O7

Hence, at last 15.72 g of dry Na2B4O7recovered

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