题目

英语翻译
A 2.50 gram of bronze was dissolved in sulfuric acid.The copper becomes copper (II) sulfate with sulfur dioxide and water the other products.The copper (II) sulfate is then mixed with iodide ion solution to form copper(I)iodide and the triiodide ion(I3-) and sulfate ions.The triiodide ion is next titrated with thiosulfate(S2O3^ 2-) to form iodide ions and S4O6^ 2- ions.Write and balance the equations and calculate the percent by mass of copper in the original sample if 21.5mL of 1.25 M thiosulfate is need in the titration.

提问时间：2020-05-08

答案

2.5g铜溶入硫酸,铜变成硫酸铜同时产生二氧化硫和水等,硫酸铜溶入碘离子溶液,形成碘化亚铜和三碘化离子和硫酸根离子.三碘化离子用硫代硫酸钠滴定从而形成碘离子和S4O6^ 2离子,写出化学方程式,并计算如果滴定反应中消耗了21.5mL 的1.25mol/L的硫代硫酸钠,消耗掉的铜的质量百分比,