assume that you start with 0.568 g of copper turnings: initial number of moles o

Question

assume that you start with 0.568 g of copper turnings:

initial number of moles of Cu : 8.94×10-3 mol

minimum volume of 16.0 M HNO3 required to dissolve all of the copper : 2.23 mL

1) 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)

2) Calculate the minimum volume of 1.00 M H2SO4 required to dissolve all of the CuO(s)

3) Calculate the minimum mass of zinc required to reduce all of the Cu2+(aq) to Cu(s)

Explanation / Answer

moles Cu = 0.568 g / 63.546 g/mol=0.00893 mole

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

moles HNO3 required = 0.00893 x 4 = 0.0357

Volume HNO3 required = 0.0357 / 16.0 M=0.00223 L => 2.23 mL

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

moles HNO3 in excess = 0.0800 - 0.0357 = 0.0443

moles NaOH required to neutralize the remaining acid = 0.0443

volume NaOH required to neutralize the remaining acid = 0.0443 / 4.00 M=0.0110 L => 11.0 mL


moles Cu(NO3)2 formed = 0.00893

moles NaOH required for the reaction = 2 x 0.00893 = 0.01786

Volume NaOH required = 0.01786/ 4.00 M=0.004465 L => 4.46 mL

volume NaOH needed to neutralize the remaining acid and convert all Cu2+ to Cu(OH)2 =11. + 4.46 =15.46 mL

moles H2SO4 required = 0.00893
V = 0.00893/ 1.00 = 0.00893 L=> 8.93 mL

moles Zn required = 0.00893
mass Zn = 0.00893 x 65.39 g/mol=0.583 g

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