Câu 4:
PTHH: \(Zn+S\underrightarrow{t^o}ZnS\)
Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\\n_S=\dfrac{0,384}{32}=0,012\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) Lưi huỳnh còn dư, Kẽm p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnS}=0,01\left(mol\right)\\n_{S\left(dư\right)}=0,002\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnS}=0,01\cdot97=0,97\left(g\right)\\m_{S\left(dư\right)}=0,002\cdot32=0,064\left(g\right)\end{matrix}\right.\)
Câu 4: (Bonus)
Ta có: \(n_{FeS}=\dfrac{17,6}{88}=0,2\left(mol\right)\)
Bảo toàn nguyên tố Lưu huỳnh: \(n_{FeS}=n_{PbS}=0,2\left(mol\right)\)
\(\Rightarrow m_{PbS}=0,2\cdot239=47,8\left(g\right)\)
