Question

A 2.000g piece of copper wire was used to precipitate silver ions as silver from a solution. When the copper metal reacts, Cu2+ ions are formed, which are soluble in water. If .750g of Ag were produced in this process, what should the final mass of the copper wire be?

Answer 1

assume solution of silver nitrate and cupper wire so reaction will be likely,

Cu + 2AgNO₃ ----> Cu(NO₃)₂ + 2Ag

we consider only Ag and Cu forget about nitrate so, we have Cu²⁺ charge on right side so reaction is correct.   

in reaction 1 mole of Cu produce 2 mole of Ag so,

moles of Ag produce = (given mass / molar mass) = (0.750 g / 107.8683 g/mol ) = 0.006952 moles

copper used to produce 0.006952 moles of Ag = (0.006952 / 2 ) = 0.003476 moles

mass of Cu = mole x molar mass = (0.003476 moles x 63.546 g/mol ) = 0.22 g

final mass of copper wire = (2.00 g - 0.22 g ) = 1.78 g