Question
I reallyyy don't understand this and I have a test on Tuesday.Please help me!! When 2.5 g of a Zinc strip were placed in a AgNO3 solution,silver metal formed on the surface of the strip. After sometime had passed, the strip was removed from the solution, dried,and weighed. If the mass of the strip was 3.37 g, calculatethe mass of Ag and Zn metals present. I reallyyy don't understand this and I have a test on Tuesday.Please help me!! When 2.5 g of a Zinc strip were placed in a AgNO3 solution,silver metal formed on the surface of the strip. After sometime had passed, the strip was removed from the solution, dried,and weighed. If the mass of the strip was 3.37 g, calculatethe mass of Ag and Zn metals present.
Explanation / Answer
I do not know if you still want the answer but here itis. When the zinc strip is placed in the silver nitrate solution,it proceeds in a substituition reaction. It is also a Redoxreaction where Zn is oxidized (0 to +2) and Ag is reduced (+1 to0) Zn + 2AgNO3 ----> 2Ag + Zn(NO3)2 Zn + 2Ag+(aq) + 2NO3-(aq) ----> Zn2+(aq) + 2NO3-(aq) + 2Ag So, when Ag is reduced to its elemental state, it deposits onthe stip in the form of a silver mirror. Moreover, byfollowing the mass law or stoichiometry, for each mole of zincoxidized, two moles of silver is reduced. Therefore, MM (Ag) = 107.9 g/mol MM (Zn) = 65.4 g/mol Change in mass = 3.37g - 2.5g = 0.87g Let moles of Zn be x Let moles of Ag be 2x So, for every mole of Zn lost to the solution, two moles of Agare added to the strip nMM = mass 107.9(2x) - 65.4x = 0.87 215.8x - 65.4x = 0.87 150.4x = 0.87 x = moles of zinc lost = 0.87/150.4 = 0.00578 mol mass of zinc lost = 0.00578 * 65.4 = 0.378 g Total Zn left on strip = 2.5-0.378 =2.12 g 2x = moles of silver added = 0.00578 * 2 = 0.0117 mol mass of Ag added onto strip = 0.0117 *107.9 = 1.25g 2x = moles of silver added = 0.00578 * 2 = 0.0117 mol mass of Ag added onto strip = 0.0117 *107.9 = 1.25g