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一、质量守恒法例1准确称取某种铁的氧化物2.88 g,用足量的CO进行还原,将生成的CO2全部用足量的澄清石灰水吸收,得到沉淀4 g,则这种铁的氧化物为( ). A.Fe2O4 B.Fe2O3 C.FeO D.以上都不是解析:据元素守恒原理可得以下关系式: CO-CO2-CaCO3 1 mol 1 mol 100 g 0.04 mol 0.04 mol 4 g 此时,气体增加的质量为0.04×(44-28)=0.64 g(即0.04 mol氧
First, the law of mass conservation 1 accurately weighed a certain kind of iron oxide 2.88 g, with a full amount of CO reduction, the generated CO2 with a full amount of clear lime water absorption, precipitation 4 g, then this iron The oxide is (). A.Fe2O4 B.Fe2O3 C.FeO D. None of the above analysis: According to the principle of element conservation, the following relationship can be obtained: CO-CO2-CaCO3 1 mol 1 mol 100 g 0.04 mol 0.04 mol 4 g The mass of gas added is 0.04×(44-28)=0.64 g (ie 0.04 mol oxygen