My friend sent me this from class and I have no clue where to start. can someone
ID: 637833 • Letter: M
Question
My friend sent me this from class and I have no clue where to start. can someone run me though this 1 to 6 please?
Prelab Questions Oxidation of Copper to Form Copper lon 1. What volume of 15.9 M concentrated nitric acid is needed to completely oxidize 0.15 g of copper? Metathesis of Copper lon to Form Copper (l) Hydroxide 2 What volume of 3.0 M sodium hydroxide is needed to completely react the copper ion generated from the oxidation of copper in question 1? How many grams of copper (I) hydroxide will be produced from the oxidation of the copper io generated in question 1? Dehydration of Copper (Il) Hydroxide to From Copper (ll) Oxide 4. If the copper (II) hydroxide from question 3 is heated to dryness, how many grams of copper (I) oxide should be produced? Decomposition of Copper (I) Oxide to Form Copper Ion 5. What volume of 6.0 M sulfuric acid is needed to react completely with the copper (I) oxide from question 4? Reduction of Copper lon to Form Copper 6, How many grams of zinc are needed to give a percent yield of 100% for the entire reaction cycle above?Explanation / Answer
Ans 1
The balanced reaction is
Cu + 2HNO3 ----> Cu(NO3)2 + H2
Moles of Cu = mass/molecular weight
= 0.15g / 63.546g/mol
= 0.00236 mol
Moles of HNO3 required
= 2 mol HNO3 x 0.00236 mol Cu / 1 mol Cu
= 0.004721 mol
Volume of HNO3 = moles/molarity
= 0.004721 mol / (15.9 mol/L)
= 0.000297 L x 1000 mL/L
= 0.297 mL
Ans 2
The balanced reaction is
Cu + 2NaOH ---> Cu(OH)2 + 2Na
Moles of Cu = 0.00236 mol
Moles of NaOH = 2 mol HNO3 x 0.00236 mol / 1 mol Cu
= 0.00472 mol
Volume of HNO3 = 0.00472 mol / (3 mol/L)
= 0.00157 L x 1000 mL/L
= 1.57 mL
Ans 3
Moles of Cu(OH)2 formed = moles of Cu consumed = 0.00236 mol
Mass of Cu(OH)2 formed = moles x molecular weight
= 0.00236 mol x 97.561 g/mol
= 0.2302 g
Ans 4
Cu(OH)2 ----> CuO + H20
Moles of CuO produced = Moles of Cu(OH)2 consumed
= 0.00236 mol
Mass of CuO produced = moles x molecular weight
= 0.00236 mol x 79.545 g/mol
= 0.1877 g
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