How much Copper (solid) can be made by reacting 0.47 moles of Sodium with 26g of Copper(II) Nitrate?2Na(s) + Cu(NO3)2 --> 2NaNO3 + Cu(s)
This is a very silly question as it is made up by someone who doesn't know their chemistry! If you mix copper nitrate crystals and sodium solid, nothing will happen unless they are left sometime. Then copper nitrate, being deliquescent will absorb water and start to dissolve. Some of the sodium will react with the water. If you try heating the solids together, copper nitrate dissolves in its water of crystallisation before decomposing and again the sodium will react partly with the water. If the copper nitrate is in solution, the sodium will react to a high percentage with the water rather than displacing metallic copper from the copper nitrate solution. 2Na(s) + 2H2O(l) ---> 2NaOH(aq) + H2(g) then Cu2+(aq) + 2OH-(aq) ---> Cu(OH)2(s) Then copper nitrate will be precipitated as copper(II) hydroxide This sort of displacement reaction works fine with other metals eg iron as iron filings, and you can do a Lab on it, but this is stupid doing a question based on misleading chemistry! At most only some of the sodium would displace copper. You can do a calculation assuming it works .. but what is the point with incorrect chemistry? What does that teach you?
what you need to do: figure out how many moles of copper nitrate you have (g*mol/g = g*1/MW = mol) figure out which of sodium and copper nitrate is the limiting reactant (ie which will get used up completed, based on the number of moles) how many moles of copper will then be produced -- if you then want your answer in grams, multiply by the molar mass
