A copper bar with a mass of 12.340 g is dipped into 255 mL of a solution containing 0.125 M silver ions. When the reaction that occurs finally ceases, what will be the mass of unreacted copper in the bar? What will be the mass of silver metal that plates out of solution (that is, what mass of silver metal is produced by the reaction)? (hint: the oxidation state of copper in aqueous solution is +2)I thought I knew how to do this but my answer doesn‘t make any sense! Please help!
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First you need an equation. Cu(s) + 2Ag+(aq) ----------- 2Ag(s) + Cu+2(aq) Convert given data into moles moles Ag+ 0.255 L x 0.125 M 0.031875 moles moles Cu(s) 17.34g / 63.55 g/mol 0.194177 moles According to the equation, for every 2 moles of Ag+ used up only half that many moles of Cu(s) will be used moles of Cu(s) used up is therefore 0.194177 / 2 0.0159375 moles Therefore moles of Cu(s) left over will be 0.194177 - 0.0159375 0.1782395 moles mass of Cu(s) unreacted will be 0.1782395 x 63.55 g/mol 11.3 g All the Ag+ ions will be consumed to produce Ag(s) So mass of Ag(s) formed will be 0.031875 x 107.9 g/mol 3.44 g
