Question

3. (14%) The metal copper is oxidized in nitric acid according to the equation: 3 Cu...

3. (14%)
The metal copper is oxidized in nitric acid according to the equation:
3 Cu + 8 HNO3 → 3 Cu (NO3) 2 + 2 NO + 4 H2O

a) Show how you set the oxidation-reduction reaction and get the equation here
above (show as many steps in the process)

b) You receive 0.3855 g of copper-containing metal together with other dissolved substances
easily set up. You find that to dissolve the copper and form
copper nitrate Cu (NO3) 2 requires 22.5 ml of 0.575 M nitric acid. What is the ratio
copper in the sample?

Show how you calculate thx :*
0 0
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Answer #1

Ans 3

Part a

The unbalanced reaction

Cu + HNO3 = Cu(NO3)2 + NO + H2O

Assign oxidation numbers for each atom within brackets

Cu(0) + H(+1)N(+5)O(-2)3 = Cu(+2)[N(+5)O(-2)3​​​​​​]2 + N(+2)O(-2) + H(+1)2O(-2)

The two half cell reactions are

Oxidation reaction

O

Cu(0) = Cu(+2)[N(+5)O(-2)3​​​​​​]2

Cu is being oxidized from 0 to +2

Reduction reaction

R:

H(+1)N(+5)O(-2)3 = N(+2)O(-2)

N is being reduced from 5 to 2

Balance the atoms except hydrogen and oxygen.

Oxidation reaction

Cu + 2HNO3 = Cu(NO3)2

Reduction reaction

HNO3 = NO

Balance the oxygen atoms by adding H2O

O:

Cu + 2HNO3 = Cu(NO3)2

R:

HNO3 = NO + 2H2O

Balance the hydrogen atoms by adding H+

O:

Cu + 2HNO3 = Cu(NO3)2 + 2H+

R:

HNO3 + 3H+ = NO + 2H2O

Balance the charge.

O:

Cu + 2HNO3 = Cu(NO3)2 + 2H+ + 2e-

R:

HNO3 + 3H+ + 3e- = NO + 2H2O

Make electron balance

O:

Cu + 2HNO3 = Cu(NO3)2 + 2H+ + 2e-

Multiply by 3

3Cu + 6HNO3 = 3Cu(NO3)2 + 6H++ 6e-

R:

HNO3 + 3H+ + 3e- = NO + 2H2O

Multiply by 2

2HNO3 + 6H+ + 6e- = 2NO + 4H2O

Add half reactions

3Cu + 8HNO3 + 6H+ + 6e- = 3Cu(NO3)2 + 2NO + 6H+ + 4H2O + 6e-

The balanced reaction

3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O

Part b

Mass of Cu = 0.3855 g

Moles of Cu = mass/molecular weight

= (0.3855g) / (63.546g/mol)

= 0.006066 mol

Volume of HNO3 = 22.5 mL x 1L/1000 mL = 0.0225 L

Molarity of HNO3 = 0.575 mol/L

Moles of HNO3 = molarity x volume

= 0.575 mol/L x 0.0225 L

= 0.0129375 mol

Mass of HNO3 = moles x molecular weight

= 0.0129375 mol x 63.01 g/mol

= 0.8152 g

Total mass of reactants = 0.3858 + 0.8152 = 1.201 g

Mass ratio of copper in the sample = mass of Cu/total mass

= 0.3858/1.201

= 0.3212

= 32.12%

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