For this net ionic equation: Cu^2+ +2OH^- --> Cu(OH)2 if 23 mL of 2.3 M copper (

Question

For this net ionic equation:   
Cu^2+ +2OH^- --> Cu(OH)2 if 23 mL of 2.3 M copper (ll) sulfate were reacted with 12 mLof 2.1 M NaOH: - What is the mass of precipitate that could form, and - What would be the concentration of all ions left insolution- Cu^2+, SO4^2-, Na^+, OH^- ?? I have no idea where to start..... thanks for all your help =] For this net ionic equation:   
Cu^2+ +2OH^- --> Cu(OH)2 if 23 mL of 2.3 M copper (ll) sulfate were reacted with 12 mLof 2.1 M NaOH: - What is the mass of precipitate that could form, and - What would be the concentration of all ions left insolution- Cu^2+, SO4^2-, Na^+, OH^- ?? I have no idea where to start..... thanks for all your help =]

Explanation / Answer

What you want to do is convert yo moles of Cu2+ and OH- moles of Cu2+=.023*2.3M=.0529 moles of Cu2+ moles of OH-=.012*2.1M=.0252 moles of OH- now convert the moles of Cu2+ and OH- to moles of Cu(OH)2 byusing the blanced equation: .0529 moles of Cu2+*1 mole Cu(OH)2/1mole of Cu2+= .0529 moles ofCu(OH)2 .0252 moles of OH-*1 mole Cu(OH)2/2moles of OH-=.0252/2 moles ofCu(OH)2 now take the smaller moles and convert it to mass of Cu(OH)2

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