1)
\(M_2S_m+\dfrac{2m+n}{2}O_2\underrightarrow{t^o}M_2O_n+mSO_2\)
\(M_2S_m+2mHCl\rightarrow2MCl_m+mH_2S\)
\(2H_2S+SO_2\rightarrow3S+2H_2O\)
\(CuSO_4+H_2S\rightarrow CuS\downarrow+H_2SO_4\)
2)
\(n_{CuS}=\dfrac{6,72}{96}=0,07\left(mol\right)\)
PTHH: \(CuSO_4+H_2S\rightarrow CuS+H_2SO_4\)
0,07<---0,07
\(n_S=\dfrac{18,24}{32}=0,57\left(mol\right)\)
PTHH: \(2H_2S+SO_2\rightarrow3S+2H_2O\)
0,38<--0,19<--0,57
- Nếu hiệu suất tính theo H2S thì \(H\%=\dfrac{0,38}{0,38+0,07}.100\%=84,44\%\)
=> Vô lý
Vậy hiệu suất tính theo SO2
\(n_{SO_2\left(bđ\right)}=\dfrac{0,19.100}{95}=0,2\left(mol\right)\)
\(n_{H_2S\left(bđ\right)}=0,38+0,07=0,45\left(mol\right)\)
- Phần 1: Gọi \(n_{M_2S_m}=a\left(mol\right)\)
Theo PTHH: \(n_{SO_2}=am=0,2\left(mol\right)\) (1)
- Phần 2: Gọi \(n_{M_2S_m}=ak\left(mol\right)\)
Theo PTHH: \(n_{H_2S}=akm=0,45\left(mol\right)\)
=> k = 2,25
Ta có: \(n_{M_2S_m\left(bđ\right)}=a+ak=\dfrac{32,5}{2.M_M+32m}\left(mol\right)\)
=> \(a=\dfrac{10}{2.M_M+32m}\left(mol\right)\)
(1) => \(\dfrac{10m}{2.M_M+32m}=0,2\)
=> MM = 9m (g/mol)
Xét m = 3 thỏa mãn => MM = 27 (g/mol)
=> M là Al
