铜镁合金4.6克完全溶于浓硝酸中,HNO3被还原产生了4.48L的NO2和0.336L的N2O4,再加入NaOH过量,求沉淀的质

4480ml=4.48L n(NO2)=0.2mol
336ml=0.336L n(N2O4)=0.015mol
由HNO3转变为NO2和N2O4过程中得到电子：n=0.2mol+0.015*2=0.23mol
建立方程：n(Cu)*64g/mol+n(Mg)*24g/mol=4.6
2*n(Cu)+2*n(Mg)=0.23mol
解出：n(Cu)=0.046mol n(Mg)=0.069mol
沉淀总质量：m=0.046mol*98g/mol+0.069mol*58g/mol
=8.51g