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When sending copper metal through a cycle of reactions, it Is most Important to

ID: 892471 • Letter: W

Question

When sending copper metal through a cycle of reactions, it Is most Important to have an idea how much of any given reagent is needed to bring each step of the cycle of reactions to completion. In order to assure a complete reaction, reagents are often added in excess (up to 10 fold), but too much as well as too little might hamper the course of a reaction. The five key steps of the copper cycle are shown below: Step 1: 3 Cu(s) + 2NO_3 + 8H^ rightarrow 3 Cu^2+(aq) + 4 H_2O + 2 NO Step 2: Cu^2+ (aq) + 2 OH (aq) rightarrow Cu(OH)_2(s) Step 3: Cu(OH)_2(s) CuO(s)+ H_2O Step 4: CuO(s) + 2 H^+ rightarrow Cu^2+(aq) + H_2O Step 5: Cu^2+(aq) + Zn(s) rightarrow Zn^2+(aq) + Cu(s) Assume that you want to carry out a sequence of cycle reactions based on 1.00 g of copper. In order to use the right amount of reagents, present answers to the following questions: What volume of 16 M HNO_3 is required to completely react with Cu in the first step of the cycle? What volume of 3.0 M NaOH is required to precipitate all copper(II) cations as Cu(OH)_2 in the second step of the cycle? How many grams of copper(II)oxide will form when the third step of the cycle goes to completion? What volume of 6.0 M H_2SO_4 is required to completely convert all copper(II)oxide to copper(II) cation in the fourth step of the cycle? How many grams of zinc metal are needed to completely regenerate all copper in the fifth step in the cycle?

Explanation / Answer

1)    MOLES OF COPPER = 1/63.5 MOLES

FOR 3 MOLES OF COPPER , 2 MOLES OF HNO3 IS NEEDED AS PER FIRST REACTION

SO FOR 1/63.5 MOLES OF COPPER   , MOLES OF HNO3 NEEDED = 2/3 ( 1/63.5) = .0105 MOLES

VOLUME OF HNO3 = .656 ML

2) FOR EACH MOLE OF CU2+ 2 MOLES OF NaOH IS NEEDED SO FOR PPT = 2 X ( 1/63.5)

3) THE MOLES OF COPPER OXIDE = ( 1/63.5)     = 79.5 X ( 1/63.5) = 1.25 GRAM

4)   THE VOLUME OF SULPHURIC ACID NEEDED = 1000/(6X 63.5 ) = 2.62 ML

5) THE MASS OF Zn NEEDE = ( 1/63.5) X 65 = 1.02 G

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