Ore floatation is the first step for copper refinery industry. In the process, c

Question

Ore floatation is the first step for copper refinery industry. In the process, copper ore sludge (2000 m3/day) was introduced into floatation pool with mean reaction time of 6 hours. Meanwhile, air was continuously and homogenously bubbled in the pool with sulfur-based organic agents. The typical O2 demand presented in flotation pool is 150 mg/liter. After reaction, the formed organic copper complex, so-called “copper concentrated” will be scrubbed out, and only clean water left.

Problem 2: Ore floatation is the first step for copper refinery industry. In the process, copper ore sludge (2000 m3/day) was introduced into floatation pool with mean reaction time of 6 hours. Meanwhile, air was continuously and homogenously bubbled in the pool with sulfur-based organic agents. The typical 02 demand presented in flotation pool is 150 mg/liter. After reaction, the formed organic copper complex, so-called "copper concentrated" will be scrubbed out, and only clean water left. (10 points) The process can be plots as: Ores sludge with volume rate of Copper concentrated Flotation pool Mean reaction time (2000 m /day) : 6 hours Clean water 150 mg O Needed/liter No Cu, No o Find the reaction rate (mole amount of reacted species per volume per time) based on O2 as limiting reactant

Explanation / Answer

since O2 is the limiting reactnat, -rO2 = Delta change in O2 concentration/ Delta change in time

in 6 hours, 150 mg/liter of O2 is needed which is equivalent to 150/32 = 4.79 mmoles/liter.

Hence, rate= 4.79/6 = 0.78125 mmoles/liter/Hr

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