How many grams of Ag will be produced from 19.0g of Cu and 125g of AgNO3?

Question

The equation is: Cu+2AgNO3 = 2Ag+Cu(NO3)2

Answer 1

Cu + 2AgNO₃ ==> 2Ag + Cu(NO₃)₂ ... balanced equation. This is both a single replacement reaction and an oxidation reduction reaction.

Moles of Cu present = 19.0 g Cu x 1 mole Cu/63.55 g = 0.2990 moles Cu

Moles AgNO3 = 125 g AgNO₃ x 1 mole AgNO₃/169.9 g = 0.7357 moles AgNO₃

Which reactant is limiting? It will be Cu because the mole ratio is 2 AgNO₃ to 1 Cu and there is more than enough AgNO₃. Thus, amount of Ag formed will depend on moles of Cu (0.2990)

Moles of Ag formed = 0.2990 moles Cu x 2 moles Ag/mole Cu = 0.598 moles Ag

Mass (grams) of Ag formed = 0.598 moles Ag x 107.9 g/mole = 64.52 g = 64.5 g of Ag (3 sig. figs.)