A water solution contains 62% Na2S202 together with 1.4% soluble impurity. The s

Question

A water solution contains 62% Na2S202 together with 1.4% soluble impurity. The solution is first diluted with water and then cooled to 15 degrees Celsius where Na2S2O2*3H2O crystalizes out. The solubility of this hydrate at 15 degrees Celsius is 1.46 lb. Na2S2O2*3H2O per lb. free water. The crystals are then separated in the first stage filter and carry an adhering solution in the proportion: 0.058 lb. solution per lb. crystals. The separated crystals are then dried in a second stage to remove the remaining free water (but not the water of hydration). If the final dry Na2S2O2*3H20 crystals must not contain more than 0.18wt% impurity, calculate... The amount of water added before cooling The %recovery of the dried crystals

Explanation / Answer

This looks like a chemical engineering problem. It will require a careful mass balance for total Na2S2O2 (as anhydrous salt and as hydrate), water (free and in crystals) and impurity. We also need an interpretation of the problem statements, and some assumptions and approximations.

Let

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