Question 5 A student decomposed 3.67g of copper (II) hydroxide into copper (II)

Question

Question 5

A student decomposed 3.67g of copper (II) hydroxide into copper (II) oxide. How many mL of 3M H2SO4is need to react with all the copper (II) oxide?

( Answer should be in decimal form to 2 decimal places)

Question 6

Using the information in question 5, how much copper (II) sulfate, in grams, are prodcued?

( Answer in decimal form to one decimal place.)

Question 7

Using the amount of copper (II) sulfate produced, in question 6, how much copper (in grams) is produced if excess zinc is used?

( Answer in decimal form to two decimal places)

Explanation / Answer

m = 3.67g Cu(OH)2

V = ??

M = 3M of H2SO4

calculate moles of Cu(OH)2

Mass of Cu(OH)2 = 97.561 g/gmol

mol of Cu(OH)2 = mass/MW = 3.67 / 97.561 = 0.037617 mols of Cu(OH)2

Cu(OH)2 + H2SO4 ----> CuSO4 + 2H2O

ratio is 1 mol of Cu(OH)2 : 1 mol of H2SO4

therefore

0.037617 mol of Cu(OH)2 will react with 0.037617 mol of H2SO4

We have n = 0.037617 mol H2SO4

M = mol/V

V = mol/M = 0.037617/3 = 0.012539 liter

or V= 12.54 ml

CuSO4 will be in solution, but if you were to precipitate, the amount will be

n= 0.037617 mol of CuSO4 were produced

MW of CuSO4 anhydrous = 159.609

mass = n*MW = 0.037617*159.609 = 6.004 grams

Question 7

CuSO4 + Zn+2 --> ZnSO4 + Cu+2

then

ratio is once again 1:1

moles of CuSO4 present = 0.037617 mol

Therefore ZnSO4 = 0.037617 mol

MW of ZnSO4 = 161.47 g/gmol

mass = mol*MW = 0.037617*161.47 = 6.07 g of ZnSO4

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.