Double Replacement Reactions

2 reactions in this lab were double replacement reactions that formed precipitates.

Part B was the first of these two. We started with Cu(NO3)2 and added NaOH. This formed a solid Cu(OH)2c and an aqueous solution NaNO3. The driving force was formation of a solid, and since NaOH is soluable, Na(NO3)2 formed a solid

The second precipitation reaction occured in Part E. We started with a solution of Cu(PO4)2 and added H2SO4. This formed aqueous NaCl and solid Cu3(PO4)2. The driving force in this reaction was formation of a solid and since NaCl is soluable, Cu3(PO4)2 formed a precipitate

Sigle Replacement Reaction

Part G was a single replacement reaction. Zinc is reacting with cupric sulfate solution to form zinc sulfate and solid copper (Zn+ Cu(SO4)=ZnSO4+Cu) The driving force in this reaction was creation of a solid (copper)

Striving for Accuracy

In picture #1, Carley is constantly stirring the solution to ensure that the reaction takes place properly

In picture #2, Lexi is down at eye level, making sure that there is exactly 100ml of solution

In picture #3, Olivia and Carley are being precise while holding the beaker up to the light to see whether the precipitation reaction is complete

In picture #4, Olivia is washing off the outside of the beaker with distilled water to ensure that all the copper solution makes it to the filter

Part G: CuSO4(aq)+Zn(s)=ZnSO4(aq)+Cu(s)

In Part F we added solid zinc to our cupric sulfate solution. These two compounds reacted and formed aqueous zinc sulfte (ZnSO4) and solid copper. We then filtered out the ZnSO4 and were left with only solid copper, exactly what we started with!

Part F: Cu3(PO4)2(s)+3H2SO4(aq)=3CuSO4(aq)+2H3PO4(aq)

In Part F we added sulfuric acid (H2SO4) to the solid cupric phosphate. This caused aqueous cupric sulfate and aqueous phosphoric acid to form and filter though to the beaker.

Part E: 3CuCl2(aq)+2Na3PO4(aq)=Cu3(PO4)2(s)+6NACl(aq)

In Part E we took the curpic chloride solution and added sodium phosphate to it. This caused a precipitate (Cu3(PO4)2) to form, as well as aqueous NaCl. The picture shows the precipitate beginning to form. We also filtered the NaCl out of the solid in this step.

Part D: CuO(s)+2HCl(aq)=CuCl2(aq)+H2O(aq)

In Part D we added hydrocloric acid (HCl) to the solid cupric oxide. This caused aqueous green cupric chloride (CuCl2) and water to form. The new soultion then drained into the beaker under the filter