n(S)==0.15mol,若SO2+2H2S=3S↓+2H2O、Cu+2H2SO4(浓)CuSO4+SO2↑+2H2O恰好反应, 由反应可知,n(SO2)=0.15mol×=0.05mol,n(H2S)=×0.15mol=0.1mol,n(Cu)=0.05mol, A.若恰好反应,a+b=0.05mol×64g/mol+0.1mol×34g/mol=6.6,故A正确; B.生成0.15molS,Cu+2H2SO4(浓)CuSO4+SO2↑+2H2O中转移电子为0.05mol×2=0.1mol,SO2+2H2S=3S↓+2H2O中转移电子为0.05mol×(4-0)=0.2mol, 即恰好完全反应时,转移电子为+=0.3mol,而SO2、H2S的反应中可能某物质过量,则共转移电子数目大于0.3mol,故B错误; C.若Cu完全反应,由电子守恒可知,0.05molCu反应,则a=0.05mol×64g/mol=3.2g,SO2+2H2S=3S↓+2H2O中硫化氢可能过量,则b>0.1mol×34g/mol=3.4g,故C正确; D.若SO2+2H2S=3S↓+2H2O中硫化氢完全反应,b=0.1mol×34g/mol=3.4g,二氧化硫过量,则a>0.05mol×64g/mol=3.2g,故D正确; 故选B. |