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Molecular and ionic equations - Assignment Example

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Molecular equation Cu(s)+4NHO3 (aq) Cu(NO3)2 (aq)+2NO2 (g)+2H2O (g) Ironic equation Cu(s)+4H+(aq)+4NO3-(aq) Cu2+(aq)+2NO3-(aq)+ 2N3+(aq)+4O2-(aq)+ 2H2O (g) Step 2 Molecular equation Cu(NO3)2 (aq)+2NaOH(aq) Cu(OH)2 (s)+2NaNO3 (aq)…
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Molecular and ionic equations
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s Post-Laboratory Assignment Molecular and ionic equations Step1 Molecular equation Cu(s)+4NHO3 (aq) Cu(NO3)2 (aq)+2NO2 (g)+2H2O (g) Ironic equation Cu(s)+4H+(aq)+4NO3-(aq) Cu2+(aq)+2NO3-(aq)+ 2N3+(aq)+4O2-(aq)+ 2H2O (g) Step 2 Molecular equation Cu(NO3)2 (aq)+2NaOH(aq) Cu(OH)2 (s)+2NaNO3 (aq) Ionic equation Cu2+(aq)+ 2NO3-(aq)+2Na+(aq)+2OH-(aq) Cu2+(aq)+ 2OH-(aq)+2 Na+(aq)+ 2NO3-(aq) Step 3 Molecular equation Cu(OH)2 (s) CuO(s) +H2O(g) Ionic equation Cu2+(aq)+ 2OH-(aq) Cu2+(aq)+O2-+ H2O(g) Step 4 Molecular equation CuO(s)+H2SO4 (aq) CuSO4(aq)+ H2O(l) Ionic equation Cu2+(aq)+O2-(aq)+2H+(aq)+SO42-(aq) Cu2+(aq)+ SO42-(aq)+ H2O(l) Step 5 Molecular equation CuSO4(aq)+Zn(s) Cu(s)+ZnSO4(aq) Ionic equation Cu2+(aq)+ )+ SO42-(aq)+Zn(s) Cu(s)+Zn2+(aq)+ SO42-(aq) Side reactions a) Molecular equation Zn(s)+HCl(aq) ZnCl2 (aq)+H2 (g) Ionic equation Zn(s)+H+(aq)+Cl-(aq) Zn2+(aq)+Cl-(aq)+H2 (g) b) Molecular equation HCl(aq)+NH3 (aq) NH4Cl (aq) Ionic equation H+(aq)+Cl-(aq) NH4+(aq) 2.

The mas of copper could be in excess as it may have oxidized to form copper (II) Oxide and also due to the side reactions that takes place. 3. i. Step 1 Calculation Number of moles in Cu is given by Mass of Cu=0.5g, Formula mass=63.546 moles Concentrated Nitric acid has a molarity of 15M, Thus there are 15moles/L of solution Number of moles in 4ml=0.004*15 =0.06moles. This implies that 0.00786 moles of Cu reacts with 0.06 moles of concentrated HNO3. Therefore, Nitric acid = excess reactant and Cu = limiting reactant. ii. Step 2 Calculation No.

of moles of 30ml NaOH of 3M concentration 3M 3moles/L, Number of moles in 30ml=0.03*3 =0.09moles of NaOH From the reaction ratio Cu(NO3)2: NaOH=1:2 But, of NaOH Moles of Cu(NO3)2 Moles This implies that 0.045moles of Cu(NO3)2 reacts with 0.09moles of NaOH, From the above it is clear Cu(NO3)2 = limiting reactant and NaOH = excess reactant Limiting reactant Excess reactant Step 1 Cu HNO3 Step 2 Cu(NO3)2 NaOH iii. Did an acid-base reaction occur? Explain Yes, in aqueous solution, H+ and OH- which reacts to form water. 4. A balanced equation for the acid base reaction that occurred. H+(aq)+ OH-(aq) H2O (l)

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