1. Assume that you start with 0.495 g of copper turnings: a)Calculate the initia

Question

1. Assume that you start with 0.495 g of copper turnings:

a)Calculate the initial number of moles of Cu.
ANSWER CALCULATED ALREADY: 7.79×10-3 mol

b) Calculate the minimum volume of 16.0 M HNO3 required to dissolve all of the copper (step 1).

c)If, in practice, 5.0 mL of 16.0 M nitric acid were added to effect the dissolution of the copper (step 1), what would be the minimum volume of 4.00 M NaOH required to neutralize the remaining acid and convert all of the Cu2+(aq) to Cu(OH)2(s) (step 2)?

d) Calculate the minimum volume of 1.00 M H2SO4 required to dissolve all of the CuO(s) (step 4).

e) Calculate the minimum mass of zinc required to reduce all of the Cu2+(aq) to Cu(s) (step 5).
(Note that, in practice, a significant excess of zinc is used due to loss of H2(g).)

Explanation / Answer

1)
a) mass of Cu=0.495g
molar mass of Cu=63.546

initial moles= mass /molar mass= 0.495/63.546= 7.79×10-3 mol

b) using M1V1=M2V2
        M=molarity, V=volume

Cu + 4 HNO3 = Cu(NO3)2 + 2 NO2 + 2 H2O

moles HNO3 required =7.79×10-3 mol x 4 = 0.0312

Volume HNO3 required = 0.0312 / 16.0 M=0.00195 L =>1.95 mL

c)
2NaOH + Cu(NO3)2 --> 2NaNO3 + Cu(OH)2

moles HNO3 added = 5.0 x 10^-3 L x 16.0 M=0.0800

moles HNO3 in excess = 0.0800 - 0.0312 = 0.0488

moles NaOH required to neutralize the remaining acid = 0.0488

volume NaOH required to neutralize the remaining acid = 0.0488 / 4.00 M=0.0122 L => 12.2 mL

moles Cu(NO3)2 formed = 7.79×10-3 mol

moles NaOH required for the reaction = 2 x 7.79×10-3 mol = 0.01558

Volume NaOH required = 0.01558/ 4.00 M=0.003895 L => 3.895 mL

volume NaOH needed to neutralize the remaining acid and convert all Cu2+ to Cu(OH)2 =12.2 - 3.895 =8.305 mL

d)
H2SO4+ CuO--> H2O + CuSO4

moles H2SO4 required = 7.79×10-3 mol
V = 7.79×10-3 mol/ 1.00 = 0.00779 L=> 7.79 mL

e)
moles Zn required = 7.79×10-3 mol
mass Zn = 7.79×10-3 mol x 65.39 g/mol=0.509 g

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