A copper strip was placed in a 250 ml silver nitrate solution with a concentration of 2 mol

A copper strip was placed in a 250 ml silver nitrate solution with a concentration of 2 mol and a density of 1.25 g / ml. After a while, the mass of the plate increased by 22.8 g. Find the mass fraction of silver nitrate in the solution at the end of the reaction.

Cu + 2AgNO3 -> Cu (NO3) 2 + 2Ag. Copper dissolves simultaneously with the release of silver. The masses of dissolved copper and released silver are related by the ratio Cu / 2Ag = 63.5 / 2 * 107.9 = 1: 3.4. Those. 1 g of dissolved copper increases 3.4 g of silver. Let X be the mass of released silver, the mass of dissolved copper then X / 3.4 = 0.3X. Then X-0.3X = 22.8, so X = 32.57g of silver was released. AgNO3 / Ag = 170 / 107.9 = 1.57, therefore 32.57 * 1.57 = 51.135 g of silver nitrate reacted. Let’s find the mass of the solution before the reaction m = V * p = 250 * 1.25 = 312.5 g. The mass of silver nitrate before the reaction is m = 170 * 2/4 = 85g, which means that 85-51.135 = 33.865g has not reacted. Copper nitrate mass m = 0.3 * 32.57 * 188/64 = 28.9 g. Mass fraction of silver nitrate at the end of the reaction w = 33.865 / (312.5-51.135 + 28.9) = 0.1167 (11.67%).