CuSO4.5H2O containing ; 3.5 percent of a soluble impurity is dissolved continuously in sufficient water and recycled mother liquor to make a saturated solution at 80^circ C . The solution is then cooled to 25^circ C , and crystals of CuSO4.5H2O are thereby obtained . These crystals carry 10 percent of their dry weight as adhering mother liquor. The crystals are then dried to zero free water (CuSO4cdot 5H2O) . The allowable impurity in the product is 0.6 weight percent. a . Calculate the weight of the water and of recycled mother liquor required per 100 kg of impure copper sulfate. b. Calculate the percentage recovery of copper sulfate, assuming that the mother liquor not recycled is discarded. The solubility of Cu SO4.5H2O at 80^circ C is

To solve this problem, we need to use the principles of solubility and stoichiometry. Let's break down the problem step by step.<br /><br />### Part (a): Calculate the weight of the water and of recycled mother liquor required per 100 kg of impure copper sulfate.<br /><br />1. **Determine the weight of pure \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) in 100 kg of impure copper sulfate:**<br /><br /> Given:<br /> - Impure copper sulfate contains 3.5% of a soluble impurity.<br /> - Therefore, the weight of pure copper sulfate in 100 kg of impure copper sulfate is:<br /> \[<br /> \text{Weight of pure CuSO}_4 = 100 \, \text{kg} \times (1 - 0.035) = 100 \, \text{kg} \times 0.965 = 96.5 \, \text{kg}<br /> \]<br /><br />2. **Calculate the weight of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) that can be obtained:**<br /><br /> The molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is:<br /> \[<br /> \text{Molar mass of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = 159.61 + 4 \times 18.02 = 249.69 \, \text{g/mol}<br /> \]<br /><br /> Therefore, the weight of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) from 96.5 kg of pure copper sulfate is:<br /> \[<br /> \text{Weight of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = \frac{96.5 \, \text{kg} \times 1000 \, \text{g/kg}}{249.69 \, \text{g/mol}} = 385.8 \, \text{kg}<br /> \]<br /><br />3. **Determine the weight of water required to dissolve the impure copper sulfate:**<br /><br /> Since the impure copper sulfate is 3.5% impure, the remaining 96.5% is copper sulfate. We need enough water to dissolve this amount at \( 80^\circ \text{C} \).<br /><br /> Let \( x \) be the weight of water required to dissolve 96.5 kg of copper sulfate at \( 80^\circ \text{C} \).<br /><br /> Using the solubility information, we can set up a proportion:<br /> \[<br /> \frac{96.5 \, \text{kg}}{x \, \text{kg}} = \frac{\text{Solubility of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \, \text{at} \, 80^\circ \text{C}}{100 \, \text{kg}}<br /> \]<br /><br /> Solving for \( x \):<br /> \[<br /> x = \frac{96.5 \, \text{kg} \times 100 \, \text{kg}}{\text{Solubility of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \, \text{at} \, 80^\circ \text{C}}<br /> \]<br /><br />4. **Determine the weight of recycled mother liquor:**<br /><br /> The crystals carry 10% of their dry weight as adhering mother liquor. Therefore, the weight of the adhering mother liquor is:<br /> \[<br /> \text{Weight of adhering mother liquor} = 0.1 \times 385.8 \, \text{kg} = 38.58 \, \text{kg}<br /> \]<br /><br /> The remaining mother liquor that needs to be recycled is:<br /> \[<br /> \text{Remaining mother liquor} = 385.8 \, \text{kg} - 38.58 \, \text{kg} = 347.22 \, \text{kg}<br /> \]<br /><br />### Part (b): Calculate the percentage recovery of copper sulfate, assuming that the mother liquor not recycled is discarded.<br /><br />1. **Determine the weight of copper sulfate in the final product:**<br /><br /> The final product is \( 385.8 \, \text{kg} \) of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \).<br /><br />2. **Calculate the weight of the final product:**<br /><br /> The weight of the final product is:<br />