a, nS = \(\frac{32}{32}\)=1mol , nCuS = \(\frac{9,6}{96}\)= 0,1mol
pt Cu + S \(\rightarrow\) CuS
(mol) 1mol \(\rightarrow\) 0,1mol
tỉ lệ \(\frac{1}{1}\) > \(\frac{0,1}{1}\) \(\rightarrow\)S dư , tính theo CuS
b, theo pt : nCu = nCuS = 0,1mol
mCu = 0,1 . 64 = 6,4 g