Cu+4HNO3(浓)==Cu(NO3)2+2NO2+2H2O (1)
3Cu+8HNO3(稀)==3Cu(NO3)2+2NO+4H2O (2)
12.8克铜=0.2mol
气体5.6升=0.25mol
设方程(1)中铜有xmol,方程(2)中铜有ymol
x+y=0.2
2x+2/3y=0.25
x=0.0875
y=0.1125
消耗的酸的物质的量是0.0875*4+0.1125*(8/3)=0.65mol
所得气体的平均相对分子质量=(2*0.0875*(16*2+14)+0.1125*(2/3)*(14+16))/0.25=41.2