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在中学化学计算中常常遇到在同一条件下的多步反应,即过量的反应物又与上一反应的生成物继续反应,且在反应中同一物质在各步反应中克当量不一样。用当量定律解决这类计算题较为困难,而最为简便的方法是用化学方程式“叠加”法,求反应物与生成物间的摩尔数比。例1、(液——液反应)在0.5 MAlCl_3溶液200ml中,缓缓加入NaOH溶液,至少要加入4NNaOH溶液多少毫升,才能使溶液恰好澄清?(不考虑盐水解) 解:AlCl_3为0.5M×0.2升=0.1摩尔设需要加入4N NaOHxml才能使溶液恰好澄清。
In the middle school chemistry calculations, a multi-step reaction under the same conditions is often encountered, that is, the excess reactants continue to react with the products of the previous reaction, and the same substance has different gram equivalents in each step of the reaction. It is difficult to solve such computational problems with the equivalent law. The simplest method is to use the chemical equation “superimposition” method to find the molar ratio between reactants and products. Example 1 (Liquid-Liquid Reaction) In 200ml of 0.5 MAlCl 3 solution, slowly add NaOH solution, at least how many milliliters of 4N NaOH solution should be added to make the solution exactly clarified?(disregarding salt solution) Solution: AlCl 3 is 0.5M × 0.2 liters = 0.1 moles requires the addition of 4N NaOH xml to properly clarify the solution.