What would happen to the ratio of copper to silver if more wire was placed in the beaker? If you let the reaction go for less time?
Depends on which Cu/Ag ratio you mean; if the one in the reaction equation, that would not change. If you mean the masses of the two in the experiment, total metallic copper in to total metallic silver out, that would change. Since the silver nitrate was the limiting reagent, the extra copper would just sit there, and the Cu/Ag ratio would go up. Inorganic reactions tend to be pretty fast [at least compared to organic ones]. So if you left it overnight that was probably way more time than required. If you gave it only 8 hours it may have made no difference. If you gave it ten minutes then it would probably have made a big difference. You would get less silver and have more copper, so Cu/Ag would rise.
Nothing will happen to the ratio because this is determined by the chemical equation. You will always use 2 moles of Ag compared to 1 of copper because copper is +2 and Silver is +1.
Cu + 2 AgNO3 ? 2 Ag + Cu(NO3)2 (a million.259 g - a million.116 g Cu) / (sixty 3.5463 g Cu/mol) x (2 mol Ag / a million mol Cu) x (107.86822 g Ag/mol) = 0.4855 g Ag in concept ( 0.475 g) / (0.4855 g) = 0.978 = ninety seven.8% yield
The only ratios that makes sense are the ratio of the amount of copper which reacts to the amount of silver that is formed, or the ratio of copper atoms that react with silver ions. Those ratios are constant and determined by the stoichiometry of the reaction. The demo is always done with an excess of copper wire. That way the silver forms on the copper, as the solution turns blue. Cu + 2Ag+ --> 2Ag(s) + Cu2+ Adding more copper wire won't change the ratio of atoms of copper and silver ions that react.
