1 / 11

Lab #10 The Steel Wool Experiment

Lab #10 The Steel Wool Experiment. Single replacement. Fe + CuSo 4(aq) FeSo 4(aq) + Cu. Vocab. . Oxidation – Involves the loss of electrons. The species that undergoes oxidation is the reducing agent.

navid
Download Presentation

Lab #10 The Steel Wool Experiment

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lab #10 The Steel Wool Experiment

  2. Single replacement • Fe + CuSo4(aq) FeSo4(aq) + Cu

  3. Vocab • . • Oxidation – Involves the loss of electrons. The species that undergoes oxidation is the reducing agent. • Reduction – Involves the gain of electrons. The species that undergoes reduction is the oxidizing agent. • Note: If one substance is oxidized then another substance is reduced.

  4. Remember – Ox is the loss of electrons; the one that looses electrons is also called the reducing agent. Ex # 1 Fe looses 2 electron Fe0(s)→Fe2++2e- Ex # 2 Zn looses 2 electrons Zn0→ Zn2+ +2e- Oxidation

  5. Remember Red. is the gain of electrons; the one that gains electrons is the oxidizing agent Ex #3 Cl1+ +1e-→ Cl Cl gains 1 electron Ex #4 Cu+25 +2e-→ Cu (s) Copper gains 2electron Reduction

  6. Fe + CuSO4(aq) Cu + FeSO4(aq) Fe0(s) + Cu+2SO4-2(aq)  Cu0(s) + Fe+2SO4-2(aq) (Blue) (Light Blue) • One mole of Fe atoms reduces one mole of Cu+2 ions to yield one mole of Cu atoms. Fe0 (s) Fe+2 (aq) + 2e- (oxidation) Cu+2 (aq) + 2e- Cu0 (s) (reduction)

  7. Mole Ratio • The molar ratio is 1:1 so 1mole Fe = 1 mole Cu • 1 mole of copper = 63.54g • 1 mole of iron = 55.85g • The formula weight then becomes • (63.54g Cu)/(55.85g Fe) = 1.14 • Note: The ratio of copper produced to iron reacted will always be their formula weight ratio.

  8. Example • 0.300 g Fe will produce how many grams of Copper? • (63.54g Cu)/(55.85g Fe) = 1.14 • (? Cu)/(0.300g Fe) = 1.14 • (? Cu) = (1.14)  (0.300g Fe) • Cu = 0.324g

  9. Procedures • Obtain and grind approximately 1.700g of CuSO4 • Add the CuSO4 to a 250mL Erlenmeyer flask. • Add 100mL of water to the flask. • Accurately determine the mass of about 0.300g of Steel Wool (Iron) • Separate the strands of Steel Wool. • Obtain and determine the mass of a filter paper. Record this value. • Using a low heat, slowly heat and swirl your solution. • Once the CuSO4 is dissolved; add your steel wool, and swirl.

  10. Procedures Cont. • Slowly pour your solution onto your filter paper which is • located in the Buchner funnel. • Rinse your flask with water and pour the water into • your funnel. • Allow your set-up to run for 5 minutes. • Using a spatula place your filter paper and copper on a watch glass. • Place watch glass into oven for 10 minutes. • Remove watch from oven and allow to cool for 5 minutes. • Determine and record the mass of your cooled product.

  11. Due next week: • Two trials; obtain data from another group for trial # 3. • Chart on page 141 • Questions 1-7 • SHOW ALL WORK FOR FULL CREDIT

More Related