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例1.铜和镁的混合物4.6 g完全溶于浓硝酸,若反应中硝酸被还原,只产生4 480 mL的NO_2气体和336 mL的N_2O_4气体(都已折算到标准状况),在反应后的溶液中加入足量的氢氧化钠溶液,生成沉淀的质量为()。A.9.02 g B.8.51 g C.8.26 g D.7.04 g解析:本题可用整体思维法进行巧解。由整体思维法可知沉淀的质量为金属Cu、Mg的质量加上OH~-的质量,由电子守恒规律知:Cu、Mg的物质的量之和为:n(Cu,Mg)×2=
Example 1. A mixture of copper and magnesium, 4.6 g, is completely soluble in concentrated nitric acid. If the nitric acid is reduced in the reaction, it produces only 4 480 mL of NO 2 and 336 mL of N 2 O 4 gas, both converted to standard conditions. After the reaction A sufficient amount of sodium hydroxide solution was added to the solution to give a precipitate of mass (). A.9.02 g B.8.51 g C.8.26 g D.7.04 g Analysis: The problem can be solved with the overall thinking method. The mass of precipitates is the mass of metal Cu and Mg plus the mass of OH ~ -. The law of conservation of electrons shows that the sum of the quantities of Cu and Mg is n (Cu, Mg) × 2 =