Purpose: To determine how much silver is contained within one gram of silver nitrate by using a single-replacement reaction.
Hypothesis: If 1 gram of silver nitrate is combined with an excess of copper, then approximately 0.6 grams of silver will be produced in the reaction.
This experiment requires silver nitrate to use as a reactant, copper to use as a reactant, distilled water, a test tube, some filter paper, and a funnel beaker.
1) Weigh out approximately 1 gram of silver nitrate. Add this to the test tube along with distilled water.
2) Cut a 30 cm piece of copper wire, coil it, and add it to the test tube.
3) Seal the test tube and wait a day.
4) Place the funnel over the beaker, and cover the funnel with the filter paper. Pour the contents of the test tube into the funnel, and spray all the silver precipitate off the wire using distilled water. Allow the funnel and beaker to drip.
5) Separate the silver from the copper wire and measure the weight of each.
Data: Before reaction: 1 g silver nitrate 3.49 g copper After reaction: 0.35 g silver 3.193 g copper
Conclusion: Again, it is possible to use stoichiometry to determine how much silver should be produced. Looking at the balanced equation
2AgNO3 + Cu ----> 2Ag + Cu(NO3)2
we can see that there is a 2:1 ratio of moles of silver nitrate to moles of copper. The molar mass of silver nitrate is about 170 moles, and the molar mass of copper is about 63.5 moles. Solving using the same techniques used in the previous post, it can be shown that about .6 grams should be produced. The experiment itself had a comparatively small percent yield, so only .35 grams were produced.