What is a flocculant? What is the purpose of adding aluminum sulfate and calcium
ID: 518105 • Letter: W
Question
What is a flocculant? What is the purpose of adding aluminum sulfate and calcium hydroxide in steps 2 and 3 of Experiment #5 (Fun with Water). (Be chemically specific. What does a flocculant actually do? How does it perform its action in purifying water.) What is activated carbon, how is it made, and why is it used for water purification? One of the steps of Experiment #6:Cu(NO_3)_2 (aq) + 2NaOH (aq) rightarrow Cu(OH)_2(s) + 2NaNO_3(aq) Suppose you started with 500 m L of5M Cu(NO_3)_2 and reacted it with 100 m L of 2M NaOH. What is the yield of Cu(OH)_2 in grams if you obtained 50% yireld? (MUST SHOW CALCULATIONS)Explanation / Answer
2) Activated carbon is a form of carbon that has numerous small pores on its surface. These pores increases the surface area of the carbon and hence activated carbon can be used as an excellent adsorbent.
Activated carbon can be produced from carbonaceous materials like peat, wood, lignite, coal, etc by heating the source material at high temperature with hot gases. Air is then introduced into the material to remove the gases creating numerous small pores on the surface.
Activated carbon has numerous small volume pores on its surfaces. These pores can adsorb impurities from water and the impurities are therefore, trapped and hence the water purified.
3) The balanced chemical equation is given
Cu(NO3)2 (aq) + 2 NaOH (aq) -------> Cu(OH)2 (s) + 2 NaNO3 (aq)
As per the balanced equation,
1 mole Cu(NO3)2 = 2 moles NaOH = 1 mole Cu(OH)2.
Moles of Cu(NO3)2 = (500 mL)*(1 L/1000 mL)*(5 mol/L) = 2.5 mole.
Moles of NaOH = (100 mL)*(1 L/1000 mL)*(2 mol/L) = 0.2 mole.
Therefore, NaOH is the limiting reactant and hence will decide the yield of the product. Therefore, yield of Cu(OH)2 = (0.2 mole NaOH)*(1 mole Cu(OH)2/2 mole NaOH) = 0.1 mole Cu(OH)2.
Molar mass of Cu(OH)2 = 97.561 g mol-1.
Therefore, mass of Cu(OH)2 theoretically possible = (0.1 mole)*(97.561 g mol-1) = 9.7561 g.
However, we have 50% yield of the product; therefore, the actual yield of Cu(OH)2 = (50/100)*(9.7561 g) = 4.87805 g 4.878 g (ans).
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